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1003. Emergency (25)-PAT甲级真题（Dijkstra算法）

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.

Input

Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (<= 500) – the number of cities (and the cities are numbered from 0 to N-1), M – the number of roads, C1 and C2 – the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2.

Output

For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather.
All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.

Sample Input

5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1

Sample Output

2 4

题目大意：n个城市m条路，每个城市有救援小组，所有的边的边权已知。给定起点和终点，求从起点到终点的最短路径条数以及最短路径上的救援小组数目之和。如果有多条就输出点权（城市救援小组数目）最大的那个～

分析：用一遍Dijkstra算法～救援小组个数相当于点权，用Dijkstra求边权最小的最短路径的条数，以及这些最短路径中点权最大的值～dis[i]表示从出发点到i结点最短路径的路径长度，num[i]表示从出发点到i结点最短路径的条数，w[i]表示从出发点到i点救援队的数目之和～当判定dis[u] + e[u][v] < dis[v]的时候，不仅仅要更新dis[v]，还要更新num[v] = num[u], w[v] = weight[v] + w[u]; 如果dis[u] + e[u][v] == dis[v]，还要更新num[v] += num[u]，而且判断一下是否权重w[v]更小，如果更小了就更新w[v] = weight[v] + w[u];

#include <iostream>
#include <algorithm>
using namespace std;
int n, m, c1, c2;
int e[510][510], weight[510], dis[510], num[510], w[510];
bool visit[510];
const int inf = 99999999;
int main() {
    scanf("%d%d%d%d", &n, &m, &c1, &c2);
    for(int i = 0; i < n; i++)
        scanf("%d", &weight[i]);
    fill(e[0], e[0] + 510 * 510, inf);
    fill(dis, dis + 510, inf);
    int a, b, c;
    for(int i = 0; i < m; i++) {
        scanf("%d%d%d", &a, &b, &c);
        e[a][b] = e[b][a] = c;
    }
    dis[c1] = 0;
    w[c1] = weight[c1];
    num[c1] = 1;
    for(int i = 0; i < n; i++) {
        int u = -1, minn = inf;
        for(int j = 0; j < n; j++) {
            if(visit[j] == false && dis[j] < minn) {
                u = j;
                minn = dis[j];
            }
        }
        if(u == -1) break;
        visit[u] = true;
        for(int v = 0; v < n; v++) {
            if(visit[v] == false && e[u][v] != inf) {
                if(dis[u] + e[u][v] < dis[v]) {
                    dis[v] = dis[u] + e[u][v];
                    num[v] = num[u];
                    w[v] = w[u] + weight[v];
                } else if(dis[u] + e[u][v] == dis[v]) {
                    num[v] = num[v] + num[u];
                    if(w[u] + weight[v] > w[v])
                        w[v] = w[u] + weight[v];
                }
            }
        }
    }
    printf("%d %d", num[c2], w[c2]);
    return 0;
}

#include <iostream>

#include <algorithm>

using namespace std;

int n, m, c1, c2;

int e[510][510], weight[510], dis[510], num[510], w[510];

bool visit[510];

const int inf = 99999999;

int main() {

scanf("%d%d%d%d", &n, &m, &c1, &c2);

for(int i = 0; i < n; i++)

scanf("%d", &weight[i]);

fill(e[0], e[0] + 510 * 510, inf);

fill(dis, dis + 510, inf);

int a, b, c;

for(int i = 0; i < m; i++) {

scanf("%d%d%d", &a, &b, &c);

e[a][b] = e[b][a] = c;

}

dis[c1] = 0;

w[c1] = weight[c1];

num[c1] = 1;

for(int i = 0; i < n; i++) {

int u = -1, minn = inf;

for(int j = 0; j < n; j++) {

if(visit[j] == false && dis[j] < minn) {

u = j;

minn = dis[j];

}

if(u == -1) break;

visit[u] = true;

for(int v = 0; v < n; v++) {

if(visit[v] == false && e[u][v] != inf) {

if(dis[u] + e[u][v] < dis[v]) {

dis[v] = dis[u] + e[u][v];

num[v] = num[u];

w[v] = w[u] + weight[v];

} else if(dis[u] + e[u][v] == dis[v]) {

num[v] = num[v] + num[u];

if(w[u] + weight[v] > w[v])

w[v] = w[u] + weight[v];

}

printf("%d %d", num[c2], w[c2]);

return 0;

}

1016. Phone Bills (25)-PAT甲级真题

A long-distance telephone company charges its customers by the following rules:

Making a long-distance call costs a certain amount per minute, depending on the time of day when the call is made. When a customer starts connecting a long-distance call, the time will be recorded, and so will be the time when the customer hangs up the phone. Every calendar month, a bill is sent to the customer for each minute called (at a rate determined by the time of day). Your job is to prepare the bills for each month, given a set of phone call records.

Input Specification:

Each input file contains one test case. Each case has two parts: the rate structure, and the phone call records.

The rate structure consists of a line with 24 non-negative integers denoting the toll (cents/minute) from 00:00 – 01:00, the toll from 01:00 – 02:00, and so on for each hour in the day.

The next line contains a positive number N (<= 1000), followed by N lines of records. Each phone call record consists of the name of the customer (string of up to 20 characters without space), the time and date (mm:dd:hh:mm), and the word “on-line” or “off-line”.

For each test case, all dates will be within a single month. Each “on-line” record is paired with the chronologically next record for the same customer provided it is an “off-line” record. Any “on-line” records that are not paired with an “off-line” record are ignored, as are “off-line” records not paired with an “on-line” record. It is guaranteed that at least one call is well paired in the input. You may assume that no two records for the same customer have the same time. Times are recorded using a 24-hour clock.

Output Specification:

For each test case, you must print a phone bill for each customer.

Bills must be printed in alphabetical order of customers’ names. For each customer, first print in a line the name of the customer and the month of the bill in the format shown by the sample. Then for each time period of a call, print in one line the beginning and ending time and date (dd:hh:mm), the lasting time (in minute) and the charge of the call. The calls must be listed in chronological order. Finally, print the total charge for the month in the format shown by the sample.

Sample Input:
10 10 10 10 10 10 20 20 20 15 15 15 15 15 15 15 20 30 20 15 15 10 10 10
10
CYLL 01:01:06:01 on-line
CYLL 01:28:16:05 off-line
CYJJ 01:01:07:00 off-line
CYLL 01:01:08:03 off-line
CYJJ 01:01:05:59 on-line
aaa 01:01:01:03 on-line
aaa 01:02:00:01 on-line
CYLL 01:28:15:41 on-line
aaa 01:05:02:24 on-line
aaa 01:04:23:59 off-line
Sample Output:
CYJJ 01
01:05:59 01:07:00 61 $12.10
Total amount: $12.10
CYLL 01
01:06:01 01:08:03 122 $24.40
28:15:41 28:16:05 24 $3.85
Total amount: $28.25
aaa 01
02:00:01 04:23:59 4318 $638.80
Total amount: $638.80

分析：将给出的数据先按照姓名排序，再按照时间的先后顺序排列，这样遍历的时候，前后两个名字相同且前面的状态为on-line后面一个的状态为off-line的就是合格数据～

注意：【关于最后一个测试点】计算费用从00:00:00到dd:hh:mm计算可以避免跨天的问题，比如01:12:00到02:02:00

#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
struct node {
    string name;
    int status, month, time, day, hour, minute;
};
bool cmp(node a, node b) {
    return a.name != b.name ? a.name < b.name : a.time < b.time;
}
double billFromZero(node call, int *rate) {
    double total = rate[call.hour] * call.minute + rate[24] * 60 * call.day;
    for (int i = 0; i < call.hour; i++)
        total += rate[i] * 60;
    return total / 100.0;
}
int main() {
    int rate[25] = {0}, n;
    for (int i = 0; i < 24; i++) {
        scanf("%d", &rate[i]);
        rate[24] += rate[i];
    }
    scanf("%d", &n);
    vector<node> data(n);
    for (int i = 0; i < n; i++) {
        cin >> data[i].name;
        scanf("%d:%d:%d:%d", &data[i].month, &data[i].day, &data[i].hour, &data[i].minute);
        string temp;
        cin >> temp;
        data[i].status = (temp == "on-line") ? 1 : 0;
        data[i].time = data[i].day * 24 * 60 + data[i].hour * 60 + data[i].minute;
    }
    sort(data.begin(), data.end(), cmp);
    map<string, vector<node> > custom;
    for (int i = 1; i < n; i++) {
        if (data[i].name == data[i - 1].name && data[i - 1].status == 1 && data[i].status == 0) {
            custom[data[i - 1].name].push_back(data[i - 1]);
            custom[data[i].name].push_back(data[i]);
        }
    }
    for (auto it : custom) {
        vector<node> temp = it.second;
        cout << it.first;
        printf(" %02d\n", temp[0].month);
        double total = 0.0;
        for (int i = 1; i < temp.size(); i += 2) {
            double t = billFromZero(temp[i], rate) - billFromZero(temp[i - 1], rate);
            printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f\n", temp[i - 1].day, temp[i - 1].hour, temp[i - 1].minute, temp[i].day, temp[i].hour, temp[i].minute, temp[i].time - temp[i - 1].time, t);
            total += t;
        }
        printf("Total amount: $%.2f\n", total);
    }
    return 0;
}

#include <iostream>

#include <map>

#include <vector>

#include <algorithm>

using namespace std;

struct node {

string name;

int status, month, time, day, hour, minute;

};

bool cmp(node a, node b) {

return a.name != b.name ? a.name < b.name : a.time < b.time;

}

double billFromZero(node call, int *rate) {

double total = rate[call.hour] * call.minute + rate[24] * 60 * call.day;

for (int i = 0; i < call.hour; i++)

total += rate[i] * 60;

return total / 100.0;

}

int main() {

int rate[25] = {0}, n;

for (int i = 0; i < 24; i++) {

scanf("%d", &rate[i]);

rate[24] += rate[i];

}

scanf("%d", &n);

vector<node> data(n);

for (int i = 0; i < n; i++) {

cin >> data[i].name;

scanf("%d:%d:%d:%d", &data[i].month, &data[i].day, &data[i].hour, &data[i].minute);

string temp;

cin >> temp;

data[i].status = (temp == "on-line") ? 1 : 0;

data[i].time = data[i].day * 24 * 60 + data[i].hour * 60 + data[i].minute;

}

sort(data.begin(), data.end(), cmp);

map<string, vector<node> > custom;

for (int i = 1; i < n; i++) {

if (data[i].name == data[i - 1].name && data[i - 1].status == 1 && data[i].status == 0) {

custom[data[i - 1].name].push_back(data[i - 1]);

custom[data[i].name].push_back(data[i]);

}

for (auto it : custom) {

vector<node> temp = it.second;

cout << it.first;

printf(" %02d\n", temp[0].month);

double total = 0.0;

for (int i = 1; i < temp.size(); i += 2) {

double t = billFromZero(temp[i], rate) - billFromZero(temp[i - 1], rate);

printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f\n", temp[i - 1].day, temp[i - 1].hour, temp[i - 1].minute, temp[i].day, temp[i].hour, temp[i].minute, temp[i].time - temp[i - 1].time, t);

total += t;

}

printf("Total amount: $%.2f\n", total);

}

return 0;

}

1021. Deepest Root (25)-PAT甲级真题（图的遍历，dfs，连通分量的个数）

A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=10000) which is the number of nodes, and hence the nodes are numbered from 1 to N. Then N-1 lines follow, each describes an edge by given the two adjacent nodes’ numbers.

Output Specification:

For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print “Error: K components” where K is the number of connected components in the graph.

Sample Input 1:
5
1 2
1 3
1 4
2 5
Sample Output 1:
3
4
5
Sample Input 2:
5
1 3
1 4
2 5
3 4
Sample Output 2:
Error: 2 components
题目大意：给出n个结点（1~n）之间的n条边，问是否能构成一棵树，如果不能构成则输出它有的连通分量个数，如果能构成一棵树，输出能构成最深的树的高度时，树的根结点。如果有多个，按照从小到大输出。
分析：首先深度优先搜索判断它有几个连通分量。如果有多个，那就输出Error: x components，如果只有一个，就两次深度优先搜索，先从一个结点dfs后保留最高高度拥有的结点们，然后从这些结点中的其中任意一个开始dfs得到最高高度的结点们，这两个结点集合的并集就是所求

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
int n, maxheight = 0;
vector<vector<int>> v;
bool visit[10010];
set<int> s;
vector<int> temp;
void dfs(int node, int height) {
    if(height > maxheight) {
        temp.clear();
        temp.push_back(node);
        maxheight = height;
    } else if(height == maxheight){
        temp.push_back(node);
    }
    visit[node] = true;
    for(int i = 0; i < v[node].size(); i++) {
        if(visit[v[node][i]] == false)
            dfs(v[node][i], height + 1);
    }
}
int main() {
    scanf("%d", &n);
    v.resize(n + 1);
    int a, b, cnt = 0, s1 = 0;
    for(int i = 0; i < n - 1; i++) {
        scanf("%d%d", &a, &b);
        v[a].push_back(b);
        v[b].push_back(a);
    }
    for(int i = 1; i <= n; i++) {
        if(visit[i] == false) {
            dfs(i, 1);
            if(i == 1) {
                if (temp.size() != 0) s1 = temp[0];
                for(int j = 0; j < temp.size(); j++)
                    s.insert(temp[j]);
            }
            cnt++;
        }
    }
    if(cnt >= 2) {
        printf("Error: %d components", cnt);
    } else {
        temp.clear();
        maxheight = 0;
        fill(visit, visit + 10010, false);
        dfs(s1, 1);
        for(int i = 0; i < temp.size(); i++)
            s.insert(temp[i]);
        for(auto it = s.begin(); it != s.end(); it++)
            printf("%d\n", *it);
    }
    return 0;
}

#include <iostream>

#include <vector>

#include <set>

#include <algorithm>

using namespace std;

int n, maxheight = 0;

vector<vector<int>> v;

bool visit[10010];

set<int> s;

vector<int> temp;

void dfs(int node, int height) {

if(height > maxheight) {

temp.clear();

temp.push_back(node);

maxheight = height;

} else if(height == maxheight){

temp.push_back(node);

}

visit[node] = true;

for(int i = 0; i < v[node].size(); i++) {

if(visit[v[node][i]] == false)

dfs(v[node][i], height + 1);

}

int main() {

scanf("%d", &n);

v.resize(n + 1);

int a, b, cnt = 0, s1 = 0;

for(int i = 0; i < n - 1; i++) {

scanf("%d%d", &a, &b);

v[a].push_back(b);

v[b].push_back(a);

}

for(int i = 1; i <= n; i++) {

if(visit[i] == false) {

dfs(i, 1);

if(i == 1) {

if (temp.size() != 0) s1 = temp[0];

for(int j = 0; j < temp.size(); j++)

s.insert(temp[j]);

}

cnt++;

}

if(cnt >= 2) {

printf("Error: %d components", cnt);

} else {

temp.clear();

maxheight = 0;

fill(visit, visit + 10010, false);

dfs(s1, 1);

for(int i = 0; i < temp.size(); i++)

s.insert(temp[i]);

for(auto it = s.begin(); it != s.end(); it++)

printf("%d\n", *it);

}

return 0;

}

1013. Battle Over Cities (25)-PAT甲级真题（图的遍历，统计强连通分量的个数，dfs）

It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of the cities connected. Given the map of cities which have all the remaining highways marked, you are supposed to tell the number of highways need to be repaired, quickly.

For example, if we have 3 cities and 2 highways connecting city1-city2 and city1-city3. Then if city1 is occupied by the enemy, we must have 1 highway repaired, that is the highway city2-city3.

Input

Each input file contains one test case. Each case starts with a line containing 3 numbers N (<1000), M and K, which are the total number of cities, the number of remaining highways, and the number of cities to be checked, respectively. Then M lines follow, each describes a highway by 2 integers, which are the numbers of the cities the highway connects. The cities are numbered from 1 to N. Finally there is a line containing K numbers, which represent the cities we concern.

Output

For each of the K cities, output in a line the number of highways need to be repaired if that city is lost.

Sample Input
3 2 3
1 2
1 3
1 2 3
Sample Output
1
0
0

题目大意：给出n个城市之间有相互连接的m条道路，当删除一个城市和其连接的道路的时候，问其他几个剩余的城市至少要添加多少个路线才能让它们重新变为连通图
分析：添加的最少的路线，就是他们的连通分量数-1，因为当a个互相分立的连通分量需要变为连通图的时候，只需要添加a-1个路线，就能让他们相连。所以这道题就是求去除了某个结点之后其他的图所拥有的连通分量数
使用邻接矩阵存储，对于每一个被占领的城市，去除这个城市结点，就是把它标记为已经访问过，这样在深度优先遍历的时候，对于所有未访问的结点进行遍历，就能求到所有的连通分量的个数～记得因为有很多个要判断的数据，每一次输入被占领的城市之前要把visit数组置为false～～

#include <cstdio>
#include <algorithm>
using namespace std;
int v[1010][1010];
bool visit[1010];
int n;
void dfs(int node) {
    visit[node] = true;
    for(int i = 1; i <= n; i++) {
        if(visit[i] == false && v[node][i] == 1)
            dfs(i);
    }
}
int main() {
    int m, k, a, b;
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 0; i < m; i++) {
        scanf("%d%d", &a, &b);
        v[a][b] = v[b][a] = 1;
    }
    for(int i = 0; i < k; i++) {
        fill(visit, visit + 1010, false);
        scanf("%d", &a);
        int cnt = 0;
        visit[a] = true;
        for(int j = 1; j <= n; j++) {
            if(visit[j] == false) {
                dfs(j);
                cnt++;
            }
        }
        printf("%d\n", cnt - 1);
    }
    return 0;
}

#include <cstdio>

#include <algorithm>

using namespace std;

int v[1010][1010];

bool visit[1010];

int n;

void dfs(int node) {

visit[node] = true;

for(int i = 1; i <= n; i++) {

if(visit[i] == false && v[node][i] == 1)

dfs(i);

}

int main() {

int m, k, a, b;

scanf("%d%d%d", &n, &m, &k);

for(int i = 0; i < m; i++) {

scanf("%d%d", &a, &b);

v[a][b] = v[b][a] = 1;

}

for(int i = 0; i < k; i++) {

fill(visit, visit + 1010, false);

scanf("%d", &a);

int cnt = 0;

visit[a] = true;

for(int j = 1; j <= n; j++) {

if(visit[j] == false) {

dfs(j);

cnt++;

}

printf("%d\n", cnt - 1);

}

return 0;

}

1076. Forwards on Weibo (30)-PAT甲级真题（图的遍历bfs）

Weibo is known as the Chinese version of Twitter. One user on Weibo may have many followers, and may follow many other users as well. Hence a social network is formed with followers relations. When a user makes a post on Weibo, all his/her followers can view and forward his/her post, which can then be forwarded again by their followers. Now given a social network, you are supposed to calculate the maximum potential amount of forwards for any specific user, assuming that only L levels of indirect followers are counted.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N (<=1000), the number of users; and L (<=6), the number of levels of indirect followers that are counted. Hence it is assumed that all the users are numbered from 1 to N. Then N lines follow, each in the format:

M[i] user_list[i]

where M[i] (<=100) is the total number of people that user[i] follows; and user_list[i] is a list of the M[i] users that are followed by user[i]. It is guaranteed that no one can follow oneself. All the numbers are separated by a space.

Then finally a positive K is given, followed by K UserID’s for query.

Output Specification:

For each UserID, you are supposed to print in one line the maximum potential amount of forwards this user can triger, assuming that everyone who can view the initial post will forward it once, and that only L levels of indirect followers are counted.

Sample Input:
7 3
3 2 3 4
0
2 5 6
2 3 1
2 3 4
1 4
1 5
2 2 6
Sample Output:
4
5

题目大意：给出每个用户关注的人的id，和转发最多的层数，求一个id发了条微博最多会有多少个人转发
分析：带层数的广度优先，因为一个用户只能转发一次，所以用inq判断当前结点是否入队过了，如果入队过了就不能重复入队（重复转发消息），inq 邻接表v 都可以使用int只存储id，queue的数据类型必须为node，同时保存它的id和layer层数，控制不超过L层～

#include <cstdio>
#include <queue>
#include <vector>
using namespace std;
int n, l, m, k;
struct node {
    int id, layer;
};
vector<vector<int>> v;
int bfs(node tnode) {
    bool inq[1010] = {false};
    queue<node> q;
    q.push(tnode);
    inq[tnode.id] = true;
    int cnt = 0;
    while(!q.empty()) {
        node top = q.front();
        q.pop();
        int topid = top.id;
        for(int i = 0; i < v[topid].size(); i++) {
            int nextid = v[topid][i];
            if(inq[nextid] == false && top.layer < l) {
                node next = {nextid, top.layer + 1};
                q.push(next);
                inq[next.id] = true;
                cnt++;
            }
        }
    }
    return cnt;
}
int main() {
    scanf("%d %d", &n, &l);
    v.resize(n + 1);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &m);
        for(int j = 0; j < m; j++) {
            int temp;
            scanf("%d", &temp);
            v[temp].push_back(i);
        }
    }
    scanf("%d", &k);
    int tid;
    for(int i = 0; i < k; i++) {
        scanf("%d", &tid);
        node tnode = {tid, 0};
        printf("%d\n", bfs(tnode));
    }
    return 0;
}

#include <cstdio>

#include <queue>

#include <vector>

using namespace std;

int n, l, m, k;

struct node {

int id, layer;

};

vector<vector<int>> v;

int bfs(node tnode) {

bool inq[1010] = {false};

queue<node> q;

q.push(tnode);

inq[tnode.id] = true;

int cnt = 0;

while(!q.empty()) {

node top = q.front();

q.pop();

int topid = top.id;

for(int i = 0; i < v[topid].size(); i++) {

int nextid = v[topid][i];

if(inq[nextid] == false && top.layer < l) {

node next = {nextid, top.layer + 1};

q.push(next);

inq[next.id] = true;

cnt++;

}

return cnt;

}

int main() {

scanf("%d %d", &n, &l);

v.resize(n + 1);

for(int i = 1; i <= n; i++) {

scanf("%d", &m);

for(int j = 0; j < m; j++) {

int temp;

scanf("%d", &temp);

v[temp].push_back(i);

}

scanf("%d", &k);

int tid;

for(int i = 0; i < k; i++) {

scanf("%d", &tid);

node tnode = {tid, 0};

printf("%d\n", bfs(tnode));

}

return 0;

}

1034. Head of a Gang (30)-PAT甲级真题（连通分量、图的遍历dfs）

One way that the police finds the head of a gang is to check people’s phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of all the phone calls made between the two persons. A “Gang” is a cluster of more than 2 persons who are related to each other with total relation weight being greater than a given threshold K. In each gang, the one with maximum total weight is the head. Now given a list of phone calls, you are supposed to find the gangs and the heads.

Input Specification:

Each input file contains one test case. For each case, the first line contains two positive numbers N and K (both less than or equal to 1000), the number of phone calls and the weight threthold, respectively. Then N lines follow, each in the following format:

Name1 Name2 Time

where Name1 and Name2 are the names of people at the two ends of the call, and Time is the length of the call. A name is a string of three capital letters chosen from A-Z. A time length is a positive integer which is no more than 1000 minutes.

Output Specification:

For each test case, first print in a line the total number of gangs. Then for each gang, print in a line the name of the head and the total number of the members. It is guaranteed that the head is unique for each gang. The output must be sorted according to the alphabetical order of the names of the heads.

Sample Input 1:
8 59
AAA BBB 10
BBB AAA 20
AAA CCC 40
DDD EEE 5
EEE DDD 70
FFF GGG 30
GGG HHH 20
HHH FFF 10
Sample Output 1:
2
AAA 3
GGG 3
Sample Input 2:
8 70
AAA BBB 10
BBB AAA 20
AAA CCC 40
DDD EEE 5
EEE DDD 70
FFF GGG 30
GGG HHH 20
HHH FFF 10
Sample Output 2:
0

题目大意：给出1000条以内的通话记录A B和权值w，和阈值k，如果一个团伙人数超过2人并且通话总权值超过k，令团伙里面的自身权值的最大值为头目，输出所有满足条件的团伙的头目，和他们团伙里面的人数
分析：总的来说是一个判断一个图的连通分量的个数，用图的遍历解决，深度优先遍历。
1.因为给的是字母，要用两个map把它们转换成数字，从1开始排列命名所有不同的人的id，存储在两个map里面，一个字符串对应id，一个id对应字符串，方便查找，正好顺便统计了总共的人数idNumber。
2.建立两个数组，weight和G，分别存储每条边的权值和每个结点的权值，因为这两个题目中都要求得后判断。
3.用传递引用的方法深度优先dfs，这样传入的参数在dfs后还能保存想要求得的值
4.遍历过一条边之后就把这条边的权值设为0（ G[u][v] = G[v][u] = 0;）防止出现回路遍历死循环

#include <iostream>
#include <map>
using namespace std;
map<string, int> stringToInt;
map<int, string> intToString;
map<string, int> ans;
int idNumber = 1, k;
int stoifunc(string s) {
    if(stringToInt[s] == 0) {
        stringToInt[s] = idNumber;
        intToString[idNumber] = s;
        return idNumber++;
    } else {
        return stringToInt[s];
    }
}
int G[2010][2010], weight[2010];
bool vis[2010];
void dfs(int u, int &head, int &numMember, int &totalweight) {
    vis[u] = true;
    numMember++;
    if(weight[u] > weight[head])
        head = u;
    for(int v = 1; v < idNumber; v++) {
        if(G[u][v] > 0) {
            totalweight += G[u][v];
            G[u][v] = G[v][u] = 0;
            if(vis[v] == false)
                dfs(v, head, numMember, totalweight);
        }
    }
}
void dfsTrave() {
    for(int i = 1; i < idNumber; i++) {
        if(vis[i] == false) {
            int head = i, numMember = 0, totalweight = 0;
            dfs(i, head, numMember, totalweight);
            if(numMember > 2 && totalweight > k)
                ans[intToString[head]] = numMember;
        }
    }
}
int main() {
    int n, w;
    cin >> n >> k;
    string s1, s2;
    for(int i = 0; i < n; i++) {
        cin >> s1 >> s2 >> w;
        int id1 = stoifunc(s1);
        int id2 = stoifunc(s2);
        weight[id1] += w;
        weight[id2] += w;
        G[id1][id2] += w;
        G[id2][id1] += w;
    }
    dfsTrave();
    cout << ans.size() << endl;
    for(auto it = ans.begin(); it != ans.end(); it++)
        cout << it->first << " " << it->second << endl;
    return 0;
}

#include <iostream>

#include <map>

using namespace std;

map<string, int> stringToInt;

map<int, string> intToString;

map<string, int> ans;

int idNumber = 1, k;

int stoifunc(string s) {

if(stringToInt[s] == 0) {

stringToInt[s] = idNumber;

intToString[idNumber] = s;

return idNumber++;

} else {

return stringToInt[s];

}

int G[2010][2010], weight[2010];

bool vis[2010];

void dfs(int u, int &head, int &numMember, int &totalweight) {

vis[u] = true;

numMember++;

if(weight[u] > weight[head])

head = u;

for(int v = 1; v < idNumber; v++) {

if(G[u][v] > 0) {

totalweight += G[u][v];

G[u][v] = G[v][u] = 0;

if(vis[v] == false)

dfs(v, head, numMember, totalweight);

}

void dfsTrave() {

for(int i = 1; i < idNumber; i++) {

if(vis[i] == false) {

int head = i, numMember = 0, totalweight = 0;

dfs(i, head, numMember, totalweight);

if(numMember > 2 && totalweight > k)

ans[intToString[head]] = numMember;

}

int main() {

int n, w;

cin >> n >> k;

string s1, s2;

for(int i = 0; i < n; i++) {

cin >> s1 >> s2 >> w;

int id1 = stoifunc(s1);

int id2 = stoifunc(s2);

weight[id1] += w;

weight[id2] += w;

G[id1][id2] += w;

G[id2][id1] += w;

}

dfsTrave();

cout << ans.size() << endl;

for(auto it = ans.begin(); it != ans.end(); it++)

cout << it->first << " " << it->second << endl;

return 0;

}