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
AC代码：
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
vector<int>s[105];
int leaf[105];///统计第几层的叶子数。
int mx = 0;
int deep;
void dfs(int deep,int step)
{
if(s[step].size()==0)
{
if(mx<deep)
mx = deep;///取最大的深度。
leaf[deep]++;
return;
}
for(int i=0; i<s[step].size(); i++)
{
dfs(deep+1,s[step][i]);
}
}
int main()
{
int n,m;
int a,num,b;
int r[105];
memset(r,0,sizeof(r));
cin>>n>>m;
for(int i=1; i<=m; i++)
{
scanf("%d%d",&a,&num);
for(int j=1; j<=num; j++)
{
scanf("%d",&b);
r[b]++;
s[a].push_back(b);
}
}
for(int i=1; i<=n; i++)
{
if(r[i]==0)
{
dfs(0,i);///用dfs求深度，顺便记录每一层的叶子数。
}
}
for(int i=0; i<=mx; i++)
{
if(i==mx)
printf("%d\n",leaf[i]);
else
printf("%d ",leaf[i]);
}
return 0;
}