1004 Counting Leaves（详细注解+30行代码）

1004 Counting Leaves

分数 30

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作者 CHEN, Yue

单位浙江大学

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

代码长度限制

16 KB

时间限制

400 ms

内存限制

64 MB

代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=103;
vector<int>v[N];
int n,m;
int cnt[N];
int max_depth=-1;
void dfs(int node,int h){//node和h分别为结点和层数
   if(v[node].size()==0){//没有孩子的结点，即叶结点
       cnt[h]++;//该层的叶子数加1
       max_depth=max(h,max_depth);//更新最大层数
       return ;
   }
   for(int i=0;i<v[node].size();i++){//深度遍历
       dfs(v[node][i],h+1);//每次层数加1
   }
}
int main(){
cin>>n>>m;
while(m--){
   int id,k;
   cin>>id>>k;
   while(k--){
       int child;
       cin>>child;
       v[id].push_back(child);//保存每个结点的孩子结点
       }
   }
   dfs(1,0);//深度遍历，记录每层叶子数，从根结点1和第0层开始遍历
   cout<<cnt[0];//输出
   for(int i=1;i<=max_depth;i++){//输出
       cout<<' '<<cnt[i];
   }
   return 0;
}