1004 Counting Leaves (30 分) 输出树每层的叶子节点数

ゝ一纸荒年。 2021-10-30 00:42 257阅读 0赞

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

题意：输出树每层的叶子节点数

一开始看题有01，02还以为要map映射，后来看了柳神的博客发现直接%d就可以了，瞬间简单了~

#include <stdio.h>
    #include <string.h>
    #include <string>
    #include <map>
    #include <vector>
    #include <queue>
    #include <math.h>
    #include <algorithm>
    #include <iostream>
    #define INF 0x3f3f3f3f
    using namespace std;
    vector<int> v[1005];
    int l[1005],maxl;
    void dfs(int root,int level)
    {
    	if(v[root].size()==0)
    	{
    		maxl=max(maxl,level);
    		l[level]++;
    		return;
    	}
    	for(int i=0;i<v[root].size();i++)
    	dfs(v[root][i],level+1);
    }
    int main()
    {
    	int n,m,i,j,id,k,x;
    	scanf("%d%d",&n,&m);
    	while(m--)
    	{
    		scanf("%d%d",&id,&k);
    		for(i=0;i<k;i++)
    		{
    			scanf("%d",&x);
    			v[id].push_back(x);
    		}		
    	}
    	memset(l,0,sizeof l);
    	maxl=-1;
    	dfs(1,0);
    	for(i=0;i<=maxl;i++)
    	{
    		if(i)printf(" ");
    		printf("%d",l[i]);
    	}
    }