Pat(A) 1094. The Largest Generation (25)

原题目：

原题链接：https://www.patest.cn/contests/pat-a-practise/1094

1094. The Largest Generation (25)

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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

给一个家族系谱的树，求人数最多的那一代。
首行为N，M，表示N个人和M个家庭。
后M行格式为，家族成员，其孩子数，孩子1，孩子2……

解题报告

树的层次遍历，用BFS，队列优化。记录每层的人员数。

代码

#include "iostream"#include "vector"#include "queue"using namespace std;int N,M;vector<vector<int>> G;int level,ans;void init(){    cin>>N>>M;    G.resize(N + 1);    int i,k,j,x;    for(i = 0; i < M; i++){        cin>>x>>k;        G[x].resize(k);        for(j = 0; j < k; j++){            cin>>G[x][j];        }    }}void bfs(){    int root = 1;    int x,i;    queue<int> que;    level = 1;    ans = 1;    int numInLevel = 0,numInNextLevel = 0,num = 0,tempLevel = 0;    que.push(root);    numInLevel ++;    numInNextLevel;    tempLevel ++;    while(!que.empty()){        x = que.front();        que.pop();        num ++;        for(i = 0; i < G[x].size(); i++){            que.push(G[x][i]);            numInNextLevel ++;        }        if(num == numInLevel){            if(numInNextLevel > ans){                level = tempLevel + 1;                ans = numInNextLevel;            }            tempLevel ++;            numInLevel = numInNextLevel;            numInNextLevel = 0;            num = 0;        }    }}int main(){    init();    bfs();    printf("%d %d",ans,level);    //system("pause");}