1004 Counting Leaves (30 分)

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 level starting 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

Sample Solution:

#include <iostream>
using namespace std;

//使用二维数组存储树，深度优先搜索遍历树，使用now_layer标记遍历层数
//也可以使用队列实现层次遍历，此处略
int nums[100]={},node[100][100]={},max_layer=0;

void dfs(int id,int now_layer){
	if (node[id][0]==0){
		nums[now_layer]++;
		if(now_layer>max_layer)max_layer=now_layer;
	}
	else{
		int i=0;
		while(node[id][i]!=0){
			dfs(node[id][i],now_layer+1);
			i++;
		}
	}
}

int main() {
    int m,n;
    cin>>m>>n;
    for(int i=0;i<n;i++){
    	int x,k;
    	cin>>x>>k;
    	for(int j=0;j<k;j++){
    		cin>>node[x][j];
		}
	}
	dfs(1,0);
	int k=0;
	while(k<max_layer){
		cout<<nums[k]<<' ';
		k++;
	}
	cout<<nums[max_layer];
    return 0;
}

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