PAT 1094 The Largest Generation[bfs][一般]

1094 The Largest Generation（25 分）

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：求树的具有最多节点的层数， 并输出是第几层。

 这是我写的。

#include <iostream>
#include <vector>
#include <map>
#include<stdio.h>
#include<cmath>
#include<queue>
using namespace std;
vector<int> tree[202];
map<int,int> mp;
int main() {
    int n,m;
    queue<int> que;
    scanf("%d %d",&n,&m);
    int id,k,temp;
    for(int i=0;i<m;i++){
        scanf("%d %d",&id,&k);
        for(int j=0;j<k;j++){
           scanf("%d",&temp);
           tree[id].push_back(temp);
        }
    }
    que.push(1);
    que.push(-1);
    mp[1]=1;
    int level=2;
    while(!que.empty()){
        int top=que.front();que.pop();
        while(top!=-1){
            for(int i=0;i<tree[top].size();i++){
                mp[level]++;
                que.push(tree[top][i]);
            }
            top=que.front();que.pop();
        }
        //是不是得两层while,每次写这个地方都有疑惑的。
        if(!que.empty()){
            level++;que.push(-1);
        }
    }
    int maxs=0;
    id=0;
    for(int i=1;i<=level;i++){
        if(mp[i]>maxs){
            maxs=mp[i];
            id=i;
        }
    }
    printf("%d %d",maxs,id);

    return 0;
}

//用-1作为标记，双层while循环进行遍历。

下列代码均来自：https://www.liuchuo.net/archives/2223

#include <cstdio>
#include <vector>
using namespace std;
vector<int> v[100];
int book[100];
void dfs(int index, int level) {
    //传参是本节点下标和节点所在层数！
    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;
}

//没看到代码之前我都不知道还可以用dfs写。以为只可以用bfs，学习了！