1004 Counting Leaves (30 point(s))

1004 Counting Leaves (30 point(s))

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0<N<100, the number of nodes in a tree, and M (<N), the number of non-leaf nodes. Then M lines follow, each in the format:

ID K ID[1] ID[2] ... ID[K]

where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 01.

The input ends with N being 0. That case must NOT be processed.

Output Specification:

For each test case, you are supposed to count those family members who have no child for every seniority levelstarting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.

Sample Input:

2 1
01 1 02

Sample Output:

0 1

题意：计算一棵树的每一个层次的叶子节点。

思路：深度优先搜索（广度优先搜索也可以），注意标记每个结点所在的层次。直到该节点没有儿子节点结束，根据该节点的层次，cnt[]自增。

BFS参考大神链接：https://blog.youkuaiyun.com/richenyunqi/article/details/79533166

#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int MAX = 107;
vector<int> graph[MAX];
int level[MAX]={0};
int cnt[MAX]={0};
int maxLevel = 0;
void dfs(int v){
	if(graph[v].size()==0){
		cnt[level[v]]++;
		return;
	}
	for(int i=0;i<graph[v].size();i++){
		int cur = graph[v][i];
		level[cur]=level[v]+1;
		maxLevel = max(maxLevel,level[cur]);
		dfs(cur);
	}
}
int main(void){
	int N,M;cin>>N>>M;
	int id,k,a;
	while(M--){
		cin>>id>>k;
		while(k--){
			cin>>a;
			graph[id].push_back(a);
		}
	}
	int root = 1;
	dfs(root);
	for(int i=0;i<=maxLevel;i++){
		if(i==0) cout<<cnt[i];
		else cout<<" "<<cnt[i];
	}
	return 0;
}