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;
}