PAT 1094. The Largest Generation (25)

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.

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

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这道题就是给一个树，要求数出哪一层的节点数最多。用BFS可以解决，代码如下： 

#include <iostream>#include <string>#include <cstring>#include <queue>#include <vector>using namespace std;typedef struct node{int id;int level;vector<int> children;}node;node family[100];int count[100];void bfs(int x){queue<node> myqueue;family[x].level=1;myqueue.push(family[x]);while(myqueue.size()){node temp=myqueue.front();myqueue.pop();count[temp.level]++;if(temp.children.size()==0)continue;for(int i=0;i<temp.children.size();i++){int target=temp.children[i];family[target].level=temp.level+1;myqueue.push(family[target]);}}}int main(void){int N,M,i;cin>>N>>M;for(i=1;i<=N;i++){family[i].id=i;family[i].children.clear();}for(i=1;i<=M;i++){node temp;int K,child;cin>>temp.id>>K;while(K--){cin>>child;temp.children.push_back(child);}family[temp.id]=temp;}memset(count,0,sizeof(count));bfs(1);int max=0,gen=0;for(int i=1;i<=100;i++){if(count[i]>max){max=count[i];gen=i;}}cout<<max<<" "<<gen;return 0;}