PAT 1094

1094. The Largest Generation (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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.

分析：简单bfs练习题

#include <iostream>#include <queue>#include <vector>#include <algorithm>using namespace std;int n,m;int level[10000]={0};int levelnum=1;vector<int> c[101];void bfs(){queue<int> q;q.push(1);level[1]=1;int last=1;int tail=1;int cnt=0;while(!q.empty()){int temp=q.front();for(int i=0;i<c[temp].size();i++){last=c[temp][i];cnt++;q.push(c[temp][i]);}if(temp==tail){levelnum++;level[levelnum]+=cnt;tail=last;cnt=0;}q.pop();}}int main(){cin>>n>>m;int root=1;for(int i=1;i<=n;i++)c[i].clear();for(int i=0;i<m;i++){int id,childnum;cin>>id>>childnum;for(int j=0;j<childnum;j++){int child;cin>>child;c[id].push_back(child);}}bfs();int max=0,temp;for(int i=1;i<=levelnum;i++)if(level[i]>max){temp=i;max=level[i];}cout<<max<<" "<<temp<<endl; return 0; }