【C++】1004. Counting Leaves (30)

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本文介绍了一种算法，用于解决家族树结构中不同层级叶节点数量的计算问题。输入包括节点总数、非叶节点数量及各非叶节点的子节点信息。通过队列实现层级遍历，统计每层的叶节点数。

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1004. Counting Leaves (30)

时间限制

400 ms

内存限制

32000 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. Your job is to count those family members who have no child.

Input

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.

Output

For each test case, you are supposed to count those family members who have no childfor 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

1.学会使用队列。

2.其实是循环中条件的判断问题啦。

3.先解决上一层次的问题，也就是计算输出上一层的结果，再计算下一层次的数目，开头错在先算下一层次，后解决上一层次。

啊啊啊

#include<iostream>
#include<queue>
#include<vector>
using namespace std;

struct ll{
	int level;
	int it;
};

int main(){
	freopen("in.txt","r",stdin);
	int N,M;
	while(scanf("%d%d",&N,&M)!=EOF){
		vector<vector<int>> t;
		queue <ll>a;
		t.resize(N);
		int i,j;
		for(i=0;i<M;i++){
			int leaf,cc;
			scanf("%d%d",&leaf,&cc);
			leaf=leaf-1;
			for(j=0;j<cc;j++){
				int temp;
				ll demo;
				scanf("%d",&temp);
				t[leaf].push_back(temp-1);
			}
		}
		/*	if(t[1].size()==0){
		cout<<"1";
		}else{
		cout<<"0";
		}*/
		ll it;
		it.it=0;
		it.level=0;
		a.push(it);
		bool flag=true;
		int count=0,nlevel=0;

		while(a.size()>0){
			ll ii=a.front();
			if(ii.level!=nlevel){
				if(flag){
					flag=false;
				}else{
					cout<<" ";
				}
				cout<<count;
				count=0;
				nlevel=ii.level;
			}
			if(ii.level==nlevel&&t[ii.it].size()==0){
				count++;		
			}

			int ccc=t[ii.it].size();
			int level=ii.level;
			for(i=0;i<ccc;i++){
				ll bb;
				bb.it=t[ii.it][i];
				bb.level=level+1;
				a.push(bb);
			}
			a.pop();
		}
		if(flag){
			flag=true;
		}else{
			cout<<" ";
		}
		cout<<count;

	}
	return 0;
}