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
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int maxn=105;
int n,m;
int par[maxn];
int vis[maxn];
int root;
vector<int>v[maxn],ans;
void traverse(int x)
{
queue<int>q;
q.push(x);
while (!q.empty())
{
int Size=q.size();
int num=0;
while (Size--)
{
int t=q.front();
q.pop();
if(!v[t].size())
{
num++;
}
else
{
for (int i=0;i<v[t].size();i++)
{
q.push(v[t][i]);
}
}
}
ans.push_back(num);
}
}
int main()
{
scanf("%d%d",&n,&m);
memset (par,-1,sizeof(par));
memset (vis,0,sizeof(vis));
for (int i=0;i<m;i++)
{
int x;
scanf("%d",&x);
vis[x]=1;
int k;
scanf("%d",&k);
for (int i=0;i<k;i++)
{
int y;
scanf("%d",&y);
vis[y]=1;
v[x].push_back(y);
par[y]=x;
}
}
for (int i=0;i<maxn;i++)
{
if(par[i]==-1&&vis[i])
{
root=i;
break;
}
}
traverse(root);
for (int i=0;i<ans.size();i++)
{
printf("%d%c",ans[i],i==ans.size()-1?'\n':' ');
}
return 0;
}