Dijkstra – liuchuo

L3-007. 天梯地图-PAT团体程序设计天梯赛GPLT

本题要求你实现一个天梯赛专属在线地图，队员输入自己学校所在地和赛场地点后，该地图应该推荐两条路线：一条是最快到达路线；一条是最短距离的路线。题目保证对任意的查询请求，地图上都至少存在一条可达路线。

输入格式：

输入在第一行给出两个正整数N（2 <= N <=500）和M，分别为地图中所有标记地点的个数和连接地点的道路条数。随后M行，每行按如下格式给出一条道路的信息：

V1 V2 one-way length time

其中V1和V2是道路的两个端点的编号（从0到N-1）；如果该道路是从V1到V2的单行线，则one-way为1，否则为0；length是道路的长度；time是通过该路所需要的时间。最后给出一对起点和终点的编号。

输出格式：

首先按下列格式输出最快到达的时间T和用节点编号表示的路线：

Time = T: 起点 => 节点1 => … => 终点

然后在下一行按下列格式输出最短距离D和用节点编号表示的路线：

Distance = D: 起点 => 节点1 => … => 终点

如果最快到达路线不唯一，则输出几条最快路线中最短的那条，题目保证这条路线是唯一的。而如果最短距离的路线不唯一，则输出途径节点数最少的那条，题目保证这条路线是唯一的。

如果这两条路线是完全一样的，则按下列格式输出：

Time = T; Distance = D: 起点 => 节点1 => … => 终点

输入样例1：
10 15
0 1 0 1 1
8 0 0 1 1
4 8 1 1 1
5 4 0 2 3
5 9 1 1 4
0 6 0 1 1
7 3 1 1 2
8 3 1 1 2
2 5 0 2 2
2 1 1 1 1
1 5 0 1 3
1 4 0 1 1
9 7 1 1 3
3 1 0 2 5
6 3 1 2 1
5 3
输出样例1：
Time = 6: 5 => 4 => 8 => 3
Distance = 3: 5 => 1 => 3
输入样例2:
7 9
0 4 1 1 1
1 6 1 3 1
2 6 1 1 1
2 5 1 2 2
3 0 0 1 1
3 1 1 3 1
3 2 1 2 1
4 5 0 2 2
6 5 1 2 1
3 5
输出样例2:
Time = 3; Distance = 4: 3 => 2 => 5
分析：用两个Dijkstra + DFS。一个求最快路径（如果相同求路径的那条），一个求最短路径（如果相同求结点数最小的那条）～～求最快路径可以直接在Dijkstra里面求前驱结点Timepre数组~~~求最短路径因为要求结点数最小的那条，所以要用dispre的二维数组存储所有结点的最短路径，然后用DFS求出满足条件的结点数最小的那条～～

注意：
1.一开始最后一个测试用例“答案错误”，后来发现是自己在求最短路径（第二个答案distance）的时候忘记了temppath每一次深搜结束后的pop_back();
2.如果直接使用DFS的话，会导致最后一个测试用例“运行超时”～

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int inf = 999999999;
int dis[510], Time[510], e[510][510], w[510][510], Timepre[510], weight[510];
bool visit[510];
vector<int> Timepath, dispath, temppath, dispre[510];
int st, fin, minnode = inf;
void dfsTimepath(int v) {
    Timepath.push_back(v);
    if(v == st) {
        return ;
    }
    dfsTimepath(Timepre[v]);
}
void dfsdispath(int v) {
    temppath.push_back(v);
    if(v == st) {
        if(temppath.size() < minnode) {
            minnode = temppath.size();
            dispath = temppath;
        }
        temppath.pop_back();
        return ;
    }
    for(int i = 0; i < dispre[v].size(); i++) {
        dfsdispath(dispre[v][i]);
    }
    temppath.pop_back();
}
int main() {
    fill(dis, dis + 510, inf);
    fill(Time, Time + 510, inf);
    fill(weight, weight + 510, inf);
    fill(e[0], e[0] + 510 * 510, inf);
    fill(w[0], w[0] + 510 * 510, inf);
    int n, m;
    scanf("%d %d", &n, &m);
    int a, b, flag, len, t;
    for(int i = 0; i < m; i++) {
        scanf("%d %d %d %d %d", &a, &b, &flag, &len, &t);
        e[a][b] = len;
        w[a][b] = t;
        if(flag != 1) {
            e[b][a] = len;
            w[b][a] = t;
        }
    }
    scanf("%d %d", &st, &fin);
    Time[st] = 0;
    for(int i = 0; i < n; i++) {
        Timepre[i] = i;
    }
    for(int i = 0; i < n; i++) {
        int u = -1, minn = inf;
        for(int j = 0; j < n; j++) {
            if(visit[j] == false && Time[j] < minn) {
                u = j;
                minn = Time[j];
            }
        }
        if(u == -1) break;
        visit[u] = true;
        for(int v = 0; v < n; v++) {
            if(visit[v] == false && w[u][v] != inf) {
                if(w[u][v] + Time[u] < Time[v]) {
                    Time[v] = w[u][v] + Time[u];
                    Timepre[v] = u;
                    weight[v] = weight[u] + e[u][v];
                } else if(w[u][v] + Time[u] == Time[v] && weight[v] > weight[u] + e[u][v]) {
                    weight[v] = weight[u] + e[u][v];
                    Timepre[v] = u;
                }
            }
        }
    }
    dfsTimepath(fin);
    fill(visit, visit + 510, false);
    dis[st] = 0;
    for(int i = 0; i < n; i++) {
        int u = -1, minn = inf;
        for(int j = 0; j < n; j++) {
            if(visit[j] == false && minn > dis[j]) {
                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(e[u][v] + dis[u] < dis[v]) {
                    dis[v] = e[u][v] + dis[u];
                    dispre[v].clear();
                    dispre[v].push_back(u);
                } else if(e[u][v] + dis[u] == dis[v]) {
                    dispre[v].push_back(u);
                }
            }
        }
    }
    dfsdispath(fin);
    printf("Time = %d", Time[fin]);
    if(dispath == Timepath) {
        printf("; Distance = %d: ", dis[fin]);
    } else {
        printf(": ");
        for(int i = Timepath.size() - 1; i >= 0; i--) {
            printf("%d", Timepath[i]);
            if(i != 0) printf(" => ");
        }
        printf("\n");
        printf("Distance = %d: ", dis[fin]);
    }
    for(int i = dispath.size() - 1; i >= 0; i--) {
        printf("%d", dispath[i]);
        if(i != 0) printf(" => ");
    }
    return 0;
}

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#include <cstdio>

#include <algorithm>

#include <vector>

using namespace std;

const int inf = 999999999;

int dis[510], Time[510], e[510][510], w[510][510], Timepre[510], weight[510];

bool visit[510];

vector<int> Timepath, dispath, temppath, dispre[510];

int st, fin, minnode = inf;

void dfsTimepath(int v) {

Timepath.push_back(v);

if(v == st) {

return ;

}

dfsTimepath(Timepre[v]);

}

void dfsdispath(int v) {

temppath.push_back(v);

if(v == st) {

if(temppath.size() < minnode) {

minnode = temppath.size();

dispath = temppath;

}

temppath.pop_back();

return ;

}

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

dfsdispath(dispre[v][i]);

}

temppath.pop_back();

}

int main() {

fill(dis, dis + 510, inf);

fill(Time, Time + 510, inf);

fill(weight, weight + 510, inf);

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

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

int n, m;

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

int a, b, flag, len, t;

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

scanf("%d %d %d %d %d", &a, &b, &flag, &len, &t);

e[a][b] = len;

w[a][b] = t;

if(flag != 1) {

e[b][a] = len;

w[b][a] = t;

}

scanf("%d %d", &st, &fin);

Time[st] = 0;

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

Timepre[i] = i;

}

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

int u = -1, minn = inf;

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

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

u = j;

minn = Time[j];

}

if(u == -1) break;

visit[u] = true;

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

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

if(w[u][v] + Time[u] < Time[v]) {

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

Timepre[v] = u;

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

} else if(w[u][v] + Time[u] == Time[v] && weight[v] > weight[u] + e[u][v]) {

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

Timepre[v] = u;

}

dfsTimepath(fin);

fill(visit, visit + 510, false);

dis[st] = 0;

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

int u = -1, minn = inf;

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

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

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

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

dispre[v].clear();

dispre[v].push_back(u);

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

dispre[v].push_back(u);

}

dfsdispath(fin);

printf("Time = %d", Time[fin]);

if(dispath == Timepath) {

printf("; Distance = %d: ", dis[fin]);

} else {

printf(": ");

for(int i = Timepath.size() - 1; i >= 0; i--) {

printf("%d", Timepath[i]);

if(i != 0) printf(" => ");

}

printf("\n");

printf("Distance = %d: ", dis[fin]);

}

for(int i = dispath.size() - 1; i >= 0; i--) {

printf("%d", dispath[i]);

if(i != 0) printf(" => ");

}

return 0;

}

1111. Online Map (30)-PAT甲级真题（Dijkstra + DFS）

Input our current position and a destination, an online map can recommend several paths. Now your job is to recommend two paths to your user: one is the shortest, and the other is the fastest. It is guaranteed that a path exists for any request.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N (2 <= N <= 500), and M, being the total number of streets intersections on a map, and the number of streets, respectively. Then M lines follow, each describes a street in the format:

V1 V2 one-way length time

where V1 and V2 are the indices (from 0 to N-1) of the two ends of the street; one-way is 1 if the street is one-way from V1 to V2, or 0 if not; length is the length of the street; and time is the time taken to pass the street.

Finally a pair of source and destination is given.

Output Specification:

For each case, first print the shortest path from the source to the destination with distance D in the format:

Distance = D: source -> v1 -> … -> destination

Then in the next line print the fastest path with total time T:

Time = T: source -> w1 -> … -> destination

In case the shortest path is not unique, output the fastest one among the shortest paths, which is guaranteed to be unique. In case the fastest path is not unique, output the one that passes through the fewest intersections, which is guaranteed to be unique.

In case the shortest and the fastest paths are identical, print them in one line in the format:

Distance = D; Time = T: source -> u1 -> … -> destination

Sample Input 1:
10 15
0 1 0 1 1
8 0 0 1 1
4 8 1 1 1
3 4 0 3 2
3 9 1 4 1
0 6 0 1 1
7 5 1 2 1
8 5 1 2 1
2 3 0 2 2
2 1 1 1 1
1 3 0 3 1
1 4 0 1 1
9 7 1 3 1
5 1 0 5 2
6 5 1 1 2
3 5
Sample Output 1:
Distance = 6: 3 -> 4 -> 8 -> 5
Time = 3: 3 -> 1 -> 5
Sample Input 2:
7 9
0 4 1 1 1
1 6 1 1 3
2 6 1 1 1
2 5 1 2 2
3 0 0 1 1
3 1 1 1 3
3 2 1 1 2
4 5 0 2 2
6 5 1 1 2
3 5
Sample Output 2:
Distance = 3; Time = 4: 3 -> 2 -> 5

题目大意：给一张地图，两个结点中既有距离也有时间，有的单行有的双向，要求根据地图推荐两条路线：一条是最快到达路线，一条是最短距离的路线。
第一行给出两个整数N和M，表示地图中地点的个数和路径的条数。接下来的M行每一行给出：道路结点编号V1 道路结点编号V2 是否单行线道路长度所需时间
要求第一行输出最快到达时间Time和路径，第二行输出最短距离Distance和路径

分析：
1.用两个Dijkstra。一个求最短路径（如果相同求时间最短的那条），一个求最快路径（如果相同求结点数最小的那条）～～
2.求最短路径,和最快路径都可以在Dijkstra里面求前驱结点dispre和，Timepre数组～
3.dispre数组更新的条件是路径更短，或者路径相等的同时时间更短。
4.求最快路径时候要多维护一个NodeNum数组，记录在时间最短的情况下，到达此节点所需的节点数量。
Time数组更新的条件是，时间更短，时间相同的时候，如果此节点能让到达次节点是数目也变小，则更新Timepre，heNodeNum数组
5.最后根据dispre 和Timepre数组递归出两条路径，比较判断，输出最终答案～
注意：如果直接使用DFS的话，会导致最后一个测试用例“运行超时”～～

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int inf = 999999999;
int dis[510], Time[510], e[510][510], w[510][510], dispre[510],Timepre[510], weight[510],NodeNum[510];
bool visit[510];
vector<int> dispath, Timepath, temppath;
int st, fin, minnode = inf;
void dfsdispath(int v) {
    dispath.push_back(v);
    if(v == st) return ;
    dfsdispath(dispre[v]);
}
void dfsTimepath(int v) {
    Timepath.push_back(v);
    if(v == st) return ;
    dfsTimepath(Timepre[v]);
}
int main() {
    fill(dis, dis + 510, inf);
    fill(Time, Time + 510, inf);
    fill(weight, weight + 510, inf);
    fill(e[0], e[0] + 510 * 510, inf);
    fill(w[0], w[0] + 510 * 510, inf);
    int n, m;
    scanf("%d %d", &n, &m);
    int a, b, flag, len, t;
    for(int i = 0; i < m; i++) {
        scanf("%d %d %d %d %d", &a, &b, &flag, &len, &t);
        e[a][b] = len;
        w[a][b] = t;
        if(flag != 1) {
            e[b][a] = len;
            w[b][a] = t;
        }
    }
    scanf("%d %d", &st, &fin);
    dis[st] = 0;
    for(int i = 0; i < n; i++)
        dispre[i] = i;
    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(e[u][v] + dis[u] < dis[v]) {
                    dis[v] = e[u][v] + dis[u];
                    dispre[v] = u;
                    weight[v] = weight[u] + w[u][v];
                } else if(e[u][v] + dis[u] == dis[v] && weight[v] > weight[u] + w[u][v]) {
                    weight[v] = weight[u] + w[u][v];
                    dispre[v] = u;
                }
            }
        }
    }
    dfsdispath(fin);
    Time[st] = 0;
    fill(visit, visit + 510, false);
    for(int i = 0; i < n; i++) {
        int u = -1, minn = inf;
        for(int j = 0; j < n; j++) {
            if(visit[j] == false && minn > Time[j]) {
                u = j;
                minn = Time[j];
            }
        }
        if(u == -1) break;
        visit[u] = true;
        for(int v = 0; v < n; v++) {
            if(visit[v] == false && w[u][v] != inf) {
                if(w[u][v] + Time[u] < Time[v]) {
                    Time[v] = w[u][v] + Time[u];
                    Timepre[v]= u;
                    NodeNum[v]=NodeNum[u]+1;
                } else if(w[u][v] + Time[u] == Time[v]&&NodeNum[u]+1<NodeNum[v]) {
                    Timepre[v]= u;
                    NodeNum[v]=NodeNum[u]+1;
                }
            }
        }
    }
    dfsTimepath(fin);
    printf("Distance = %d", dis[fin]);
    if(dispath == Timepath) {
        printf("; Time = %d: ", Time[fin]);
    } else {
        printf(": ");
        for(int i = dispath.size() - 1; i >= 0; i--) {
            printf("%d", dispath[i]);
            if(i != 0) printf(" -> ");
        }
        printf("\nTime = %d: ", Time[fin]);
    }
    for(int i = Timepath.size() - 1; i >= 0; i--) {
        printf("%d", Timepath[i]);
        if(i != 0) printf(" -> ");
    }
    return 0;
}

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#include <iostream>

#include <algorithm>

#include <vector>

using namespace std;

const int inf = 999999999;

int dis[510], Time[510], e[510][510], w[510][510], dispre[510],Timepre[510], weight[510],NodeNum[510];

bool visit[510];

vector<int> dispath, Timepath, temppath;

int st, fin, minnode = inf;

void dfsdispath(int v) {

dispath.push_back(v);

if(v == st) return ;

dfsdispath(dispre[v]);

}

void dfsTimepath(int v) {

Timepath.push_back(v);

if(v == st) return ;

dfsTimepath(Timepre[v]);

}

int main() {

fill(dis, dis + 510, inf);

fill(Time, Time + 510, inf);

fill(weight, weight + 510, inf);

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

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

int n, m;

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

int a, b, flag, len, t;

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

scanf("%d %d %d %d %d", &a, &b, &flag, &len, &t);

e[a][b] = len;

w[a][b] = t;

if(flag != 1) {

e[b][a] = len;

w[b][a] = t;

}

scanf("%d %d", &st, &fin);

dis[st] = 0;

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

dispre[i] = i;

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

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

dispre[v] = u;

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

} else if(e[u][v] + dis[u] == dis[v] && weight[v] > weight[u] + w[u][v]) {

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

dispre[v] = u;

}

dfsdispath(fin);

Time[st] = 0;

fill(visit, visit + 510, false);

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

int u = -1, minn = inf;

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

if(visit[j] == false && minn > Time[j]) {

u = j;

minn = Time[j];

}

if(u == -1) break;

visit[u] = true;

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

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

if(w[u][v] + Time[u] < Time[v]) {

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

Timepre[v]= u;

NodeNum[v]=NodeNum[u]+1;

} else if(w[u][v] + Time[u] == Time[v]&&NodeNum[u]+1<NodeNum[v]) {

Timepre[v]= u;

NodeNum[v]=NodeNum[u]+1;

}

dfsTimepath(fin);

printf("Distance = %d", dis[fin]);

if(dispath == Timepath) {

printf("; Time = %d: ", Time[fin]);

} else {

printf(": ");

for(int i = dispath.size() - 1; i >= 0; i--) {

printf("%d", dispath[i]);

if(i != 0) printf(" -> ");

}

printf("\nTime = %d: ", Time[fin]);

}

for(int i = Timepath.size() - 1; i >= 0; i--) {

printf("%d", Timepath[i]);

if(i != 0) printf(" -> ");

}

return 0;

}

1087. All Roads Lead to Rome (30)-PAT甲级真题（Dijkstra + DFS）

Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness.

Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers N (2<=N<=200), the number of cities, and K, the total number of routes between pairs of cities; followed by the name of the starting city. The next N-1 lines each gives the name of a city and an integer that represents the happiness one can gain from that city, except the starting city. Then K lines follow, each describes a route between two cities in the format “City1 City2 Cost”. Here the name of a city is a string of 3 capital English letters, and the destination is always ROM which represents Rome.

Output Specification:
For each test case, we are supposed to find the route with the least cost. If such a route is not unique, the one with the maximum happiness will be recommended. If such a route is still not unique, then we output the one with the maximum average happiness — it is guaranteed by the judge that such a solution exists and is unique.
Hence in the first line of output, you must print 4 numbers: the number of different routes with the least cost, the cost, the happiness, and the average happiness (take the integer part only) of the recommended route. Then in the next line, you are supposed to print the route in the format “City1->City2->…->ROM”.

Sample Input:
6 7 HZH
ROM 100
PKN 40
GDN 55
PRS 95
BLN 80
ROM GDN 1
BLN ROM 1
HZH PKN 1
PRS ROM 2
BLN HZH 2
PKN GDN 1
HZH PRS 1
Sample Output:
3 3 195 97
HZH->PRS->ROM

题目大意：有N个城市，M条无向边，从某个给定的起始城市出发，前往名为ROM的城市。每个城市（除了起始城市）都有一个点权（称为幸福值），和边权（每条边所需的花费）。求从起点到ROM所需要的最少花费，并输出其路径。如果路径有多条，给出幸福值最大的那条。如果仍然不唯一，选择路径上的城市平均幸福值最大的那条路径
分析：Dijkstra+DFS。Dijkstra算出所有最短路径的路线pre二维数组，DFS求最大happiness的路径path，并求出路径条数，最大happiness，最大average～
注意：average是除去起点的average。城市名称可以用m存储字符串所对应的数字，用m2存储数字对应的字符串

#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
int n, k;
const int inf = 999999999;
int e[205][205], weight[205], dis[205];
bool visit[205];
vector<int> pre[205], temppath, path;
map<string, int> m;
map<int, string> m2;
int maxvalue = 0, mindepth, cntpath = 0;
double maxavg;
void dfs(int v) {
    temppath.push_back(v);
    if(v == 1) {
        int value = 0;
        for(int i = 0; i < temppath.size(); i++) {
            value += weight[temppath[i]];
        }
        double tempavg = 1.0 * value / (temppath.size() - 1);
        if(value > maxvalue) {
            maxvalue = value;
            maxavg = tempavg;
            path = temppath;
        } else if(value == maxvalue && tempavg > maxavg) {
            maxavg = tempavg;
            path = temppath;
        }
        temppath.pop_back();
        cntpath++;
        return ;
    }
    for(int i = 0; i < pre[v].size(); i++) {
        dfs(pre[v][i]);
    }
    temppath.pop_back();
}
int main() {
    fill(e[0], e[0] + 205 * 205, inf);
    fill(dis, dis + 205, inf);
    scanf("%d %d", &n, &k);
    string s;
    cin >> s;
    m[s] = 1;
    m2[1] = s;
    for(int i = 1; i < n; i++) {
        cin >> s >> weight[i+1];
        m[s] = i+1;
        m2[i+1] = s;
    }
    string sa, sb;
    int temp;
    for(int i = 0; i < k; i++) {
        cin >> sa >> sb >> temp;
        e[m[sa]][m[sb]] = temp;
        e[m[sb]][m[sa]] = temp;
    }
    dis[1] = 0;
    for(int i = 0; i < n; i++) {
        int u = -1, minn = inf;
        for(int j = 1; 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 = 1; 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];
                    pre[v].clear();
                    pre[v].push_back(u);
                } else if(dis[v] == dis[u] + e[u][v]) {
                    pre[v].push_back(u);
                }
            }
        }
    }
    int rom = m["ROM"];
    dfs(rom);
    printf("%d %d %d %d\n", cntpath, dis[rom], maxvalue, (int)maxavg);
    for(int i = path.size() - 1; i >= 1; i--) {
        cout << m2[path[i]] << "->";
    }
    cout << "ROM";
    return 0;
}

#include <iostream>

#include <map>

#include <vector>

#include <algorithm>

using namespace std;

int n, k;

const int inf = 999999999;

int e[205][205], weight[205], dis[205];

bool visit[205];

vector<int> pre[205], temppath, path;

map<string, int> m;

map<int, string> m2;

int maxvalue = 0, mindepth, cntpath = 0;

double maxavg;

void dfs(int v) {

temppath.push_back(v);

if(v == 1) {

int value = 0;

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

value += weight[temppath[i]];

}

double tempavg = 1.0 * value / (temppath.size() - 1);

if(value > maxvalue) {

maxvalue = value;

maxavg = tempavg;

path = temppath;

} else if(value == maxvalue && tempavg > maxavg) {

maxavg = tempavg;

path = temppath;

}

temppath.pop_back();

cntpath++;

return ;

}

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

dfs(pre[v][i]);

}

temppath.pop_back();

}

int main() {

fill(e[0], e[0] + 205 * 205, inf);

fill(dis, dis + 205, inf);

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

string s;

cin >> s;

m[s] = 1;

m2[1] = s;

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

cin >> s >> weight[i+1];

m[s] = i+1;

m2[i+1] = s;

}

string sa, sb;

int temp;

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

cin >> sa >> sb >> temp;

e[m[sa]][m[sb]] = temp;

e[m[sb]][m[sa]] = temp;

}

dis[1] = 0;

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

int u = -1, minn = inf;

for(int j = 1; 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 = 1; 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];

pre[v].clear();

pre[v].push_back(u);

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

pre[v].push_back(u);

}

int rom = m["ROM"];

dfs(rom);

printf("%d %d %d %d\n", cntpath, dis[rom], maxvalue, (int)maxavg);

for(int i = path.size() - 1; i >= 1; i--) {

cout << m2[path[i]] << "->";

}

cout << "ROM";

return 0;

}