1094. The Largest Generation (25)-PAT甲级真题（bfs,dfs,树的遍历）

A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level belong to the same generation. Your task is to find the generation with the largest population.

Input Specification:

Each input file contains one test case. Each case starts with two positive integers N (<100) which is the total number of family members in the tree (and hence assume that all the members are numbered from 01 to N), and M (<N) which is the number of family members who have children. Then M lines follow, each contains the information of a family member in the following format:

ID K ID[1] ID[2] … ID[K]

where ID is a two-digit number representing a family member, K (>0) is the number of his/her children, followed by a sequence of two-digit ID’s of his/her children. For the sake of simplicity, let us fix the root ID to be 01. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the largest population number and the level of the corresponding generation. It is assumed that such a generation is unique, and the root level is defined to be 1.

Sample Input:
23 13
21 1 23
01 4 03 02 04 05
03 3 06 07 08
06 2 12 13
13 1 21
08 2 15 16
02 2 09 10
11 2 19 20
17 1 22
05 1 11
07 1 14
09 1 17
10 1 18
Sample Output:
9 4
题目大意：输入树的结点个数N，结点编号为1~N，非叶子结点个数M，然后输出M个非叶子结点格子的孩子结点的编号，求结点个数最多的一层，根结点的层号为1，输出该层的结点个数以及层号<(▰˘◡˘▰)>
分析：用DFS或者BFS，用DFS就用参数index和level标记当前遍历的结点的编号和层数，一个数组book标记当前层数level所含结点数，最后遍历一遍数组找出最大值。注意：book[level]++;这句话是发生在return语句判断之前的外面，即每遇到一个结点都要进行处理，而不是放在return语句的条件判断里面～～
如果是BFS，就用一个数组level[i]标记i结点所处的层数，它等于它的父亲结点的level的值+1，用一个数组book，book[i]标记i层所拥有的结点数，在遍历的时候每弹出一个结点就将当前结点的层数所对应的book值+1，最后遍历一遍book数组找出最大拥有的结点数和层数～

深度优先dfs方法：

#include <cstdio>
#include <vector>
using namespace std;
vector<int> v[100];
int book[100];
void dfs(int index, int level) {
    book[level]++;
    for(int i = 0; i < v[index].size(); i++)
        dfs(v[index][i], level+1);
}
int main() {
    int n, m, a, k, c;
    scanf("%d %d", &n, &m);
    for(int i = 0; i < m; i++) {
        scanf("%d %d",&a, &k);
        for(int j = 0; j < k; j++) {
            scanf("%d", &c);
            v[a].push_back(c);
        }
    }
    dfs(1, 1);
    int maxnum = 0, maxlevel = 1;
    for(int i = 1; i < 100; i++) {
        if(book[i] > maxnum) {
            maxnum = book[i];
            maxlevel = i;
        }
    }
    printf("%d %d", maxnum, maxlevel);
    return 0;
}

#include <cstdio>

#include <vector>

using namespace std;

vector<int> v[100];

int book[100];

void dfs(int index, int level) {

book[level]++;

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

dfs(v[index][i], level+1);

}

int main() {

int n, m, a, k, c;

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

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

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

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

scanf("%d", &c);

v[a].push_back(c);

}

dfs(1, 1);

int maxnum = 0, maxlevel = 1;

for(int i = 1; i < 100; i++) {

if(book[i] > maxnum) {

maxnum = book[i];

maxlevel = i;

}

printf("%d %d", maxnum, maxlevel);

return 0;

}

广度优先bfs方法：

#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
vector<int> v[100];
int level[100];
int book[100];
int main() {
    int n, m, a, k, c;
    scanf("%d %d", &n, &m);
    for(int i = 0; i < m; i++) {
        scanf("%d %d",&a, &k);
        for(int j = 0; j < k; j++) {
            scanf("%d", &c);
            v[a].push_back(c);
        }
    }
    queue<int> q;
    q.push(1);
    level[1] = 1;
    while(!q.empty()) {
        int index = q.front();
        q.pop();
        book[level[index]]++;
        for(int i = 0; i < v[index].size(); i++) {
            level[v[index][i]] = level[index] + 1;
            q.push(v[index][i]);
        }
        
    }
    int maxnum = 0, maxlevel = 1;
    for(int i = 1; i < 100; i++) {
        if(book[i] > maxnum) {
            maxnum = book[i];
            maxlevel = i;
        }
    }
    printf("%d %d", maxnum, maxlevel);
    return 0;
}

#include <cstdio>

#include <vector>

#include <queue>

using namespace std;

vector<int> v[100];

int level[100];

int book[100];

int main() {

int n, m, a, k, c;

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

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

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

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

scanf("%d", &c);

v[a].push_back(c);

}

queue<int> q;

q.push(1);

level[1] = 1;

while(!q.empty()) {

int index = q.front();

q.pop();

book[level[index]]++;

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

level[v[index][i]] = level[index] + 1;

q.push(v[index][i]);

}