pat1004

1004. Counting Leaves (30)

时间限制

400 ms

内存限制

65536 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 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

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提交代码

题意就是 给你 2个数。第一个数就是n个点，m就是接下来m行，

每行出入 a，x 

a 这个点下面有 x个数 分别为 ......

输出每一层没有孩子节点的个数。.

这题用vector去存树==

不懂 看下这个http://www.cnblogs.com/2014slx/p/7325328.html

写一个bfs。

#include<iostream>
#include<vector>
#include<cstdio>
#include<string.h>
using namespace std;
int cnt[101];
vector<int >mp[10000];
void bfs(int q,int w)
{
    if(mp[q].empty()==1)
    {
        cnt[w]++;
        return;
    }
    for(vector <int>::iterator it=mp[q].begin();it!=mp[q].end();it++)
    {
       bfs(*it ,w+1);

    }

}


int main()
{
    memset(cnt,0,sizeof cnt);
    int n,m;
    scanf("%d%d",&n,&m);
    int x=n-m;
    while(m--)
    {
        int a,b,c;
        scanf("%d%d",&a,&b);
        for(int i=0;i<b;i++)
        {
            scanf("%d",&c);


            mp[a].push_back(c);
        }
    }
    bfs(1,0);
    printf("%d",cnt[0]);
    int num=cnt[0];
    for(int i=1;num<x;i++)
    {
        printf(" %d",cnt[i]);
        num+=cnt[i];
    }
    printf("\n");
    return 0;
}

最后一部判断就是。n-m就是 下面没有点的值。加起来等于这个就要退出。因为我们不知道层数。