Tree of Life(easy) Codeforces 690F1

http://codeforces.com/problemset/problem/690/F1

F1. Tree of Life (easy)

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Heidi has finally found the mythical Tree of Life – a legendary combinatorial structure which is said to contain a prophecy crucially needed to defeat the undead armies.

On the surface, the Tree of Life is just a regular undirected tree well-known from computer science. This means that it is a collection of npoints (called vertices), some of which are connected using n - 1 line segments (edges) so that each pair of vertices is connected by apath (a sequence of one or more edges).

To decipher the prophecy, Heidi needs to perform a number of steps. The first is counting the number of lifelines in the tree – these are paths of length 2, i.e., consisting of two edges. Help her!

Input

The first line of the input contains a single integer n – the number of vertices in the tree (1 ≤ n ≤ 10000). The vertices are labeled with the numbers from 1 to n. Then n - 1 lines follow, each describing one edge using two space-separated numbers a b – the labels of the vertices connected by the edge (1 ≤ a < b ≤ n). It is guaranteed that the input represents a tree.

Output

Print one integer – the number of lifelines in the tree.

Examples

input

4
1 2
1 3
1 4

output

3

input

5
1 2
2 3
3 4
3 5

output

4

Note

In the second sample, there are four lifelines: paths between vertices 1 and 3, 2 and 4, 2 and 5, and 4 and 5.

题目的大致意思是给你一棵树，求长度为2的路的个数(两点之间的路长度为1)

由于没有认真读题，看成了求多少组父子关系，结果WA╮(╯_╰)╭。搜了别人的题解才发现自己没有认真读题。

解法就是dfs，深度为2就+1。树其实也是图，顶点之间的关系用邻接表，为了方便，选用二维vector。要注意的是，所求结果为ans/2，因为从每一个点深度优先搜索，每条路径会重复一次(很容易理解)

AC代码如下:

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
const int maxn=1e4+5;
vector<int> G[maxn];
int visited[maxn];
int n;
int ans;
void dfs(int u,int deep)
{
    visited[u]=1;
    if(deep==2){
        ans++;
        return;
    }
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i];
        if(!visited[v]){
            dfs(v,deep+1);
        }
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<n;i++){
        int a,b;
        scanf("%d%d",&a,&b);
        G[a].push_back(b);
        G[b].push_back(a);
    }
    ans=0;
    for(int i=1;i<=n;i++){
        memset(visited,0,sizeof(visited));
        dfs(i,0);
    }
    printf("%d",ans/2);
    return 0;
}