A family hierarchy（等级制度） is usually presented（展现） by a pedigree（族谱） 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（顺序） of two-digit ID's of its children. For the sake（理由） of simplicity（简单）, 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（对于每一层） 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 1AC代码
#include <iostream>
#include <vector>
#include <map>
using namespace std;
vector<vector<int>> tree;
map<int,int>cnt;
void dfs(int cur,int depth)
{
if(!cnt[depth])
cnt[depth]=0;
if(tree[cur].empty())
{
cnt[depth]++;
return;
}
for(int i=0;i<tree[cur].size();i++)
dfs(tree[cur][i],depth+1);
}
int main()
{
int n,m,temp1,temp2,k;
cin>>n>>m;
tree.resize(n+1);
for(int i=0;i<m;i++)
{
cin>>temp1>>k;
for(int j=0;j<k;j++)
{
cin>>temp2;
tree[temp1].push_back(temp2);
}
}
dfs(1,1);
map<int,int>::iterator it;
for(it=cnt.begin();it!=cnt.end();it++)
{
if(it!=cnt.begin())
cout<<" ";
cout<<it->second;
}
return 0;
}
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