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

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

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]);
    }
}