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<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：给一棵树，数每一层叶子节点的数量，最多100个节点，n代表总共n个节点，m代表mge非叶子节点，m是2位数，1表示为01，k表示这个节点有几个孩子。
解题思路：用vector来存储，输入的m行中每一行都在同一层，通过这个来解决这题就显得比较简单了，通过v[a[front]].size()==0来判断是不是叶子节点，用数组a储存每一层的孩子，维持一下判断的顺序就好了。
#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
vector<int>v[100];
int a[100];
int main(void)
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
int x,y;
scanf("%d%d",&x,&y);
for(int j=0;j<y;j++)
{
int z;
scanf("%d",&z);
v[x].push_back(z);
}
}
int first=1;
a[0]=1;
int front=0,rear=1;
while(front<rear)
{
int tag=rear;
int cnt=0;
while(front<tag)
{
int i=0;
if(v[a[front]].size()==0) cnt++;//判断叶子节点
else
{
while(v[a[front]][i])
{
a[rear++]=v[a[front]][i];
i++;
}
}
//cout<<endl;
//cout<<front<<" "<<tag<<" "<<rear<<endl;
front++;
}
if(first) first=0;//控制输出格式
else cout<<" ";
cout<<cnt;
}
cout<<endl;
return 0;
}