PAT-1004 Counting Leaves

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

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

 

#include
#include
#include
using namespace std;

vector tree[105];
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=0;i q;
	q.push(1);
	int begin=1,leaf=0;bool pre=false,first=true;
	while(!q.empty()){
		int t=q.front();
		q.pop();
		if(t==begin){
			if(t!=1){
				if(!first)printf(" ");
				else first=false;
				printf("%d",leaf);
			}leaf=0;pre=true;
		}
		if(tree[t].size()==0)
			leaf++;
		for(int i=0;i