pat-a 1094. The Largest Generation (25)

直接广搜判断哪一层结点最多即可

#include<cstdio>#include<map>#include<vector>#include<queue>using namespace std;map<int,vector<int> >tree;int c[110];int maxn=0,p=0;struct node{int num;int level;node(int a=0,int b=0):num(a),level(b){}};void bfs(){queue<node> q;q.push(node(1,0));while(!q.empty()){node temp=q.front();int n=temp.num;int level=temp.level;q.pop();c[level]++;if(c[level]>maxn){maxn=c[level];p=level;}int len=tree[n].size();for(int i=0;i<len;++i){int k=tree[n][i];q.push(node(k,temp.level+1));}}}int main(){int n,m,a,b,t;scanf("%d%d",&n,&m);while(m--){scanf("%d%d",&a,&b);for(int i=0;i<b;++i){scanf("%d",&t);tree[a].push_back(t);}}bfs();printf("%d %d\n",maxn,p+1);return 0;} 

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 1321 1 2301 4 03 02 04 0503 3 06 07 0806 2 12 1313 1 2108 2 15 1602 2 09 1011 2 19 2017 1 2205 1 1107 1 1409 1 1710 1 18

Sample Output:

9 4