ZJH and Monkeys

链接： https://www.nowcoder.com/acm/contest/106/A
来源：牛客网

题目描述

It’s universally acknowledged that there’re innumerable trees in the campus of HUST.  

One day, zjh is wandering in the campus,and he finds a tree with n monkeys on it, and he labels the monkeys from 1 to n. A triple of monkeys (a,b,c) is called happy if and only if the absolute value of the difference of the distance from c to a the distance from c to b holds the same parity(奇偶性) as n. Now zjh wants to know how many triples of monkeys are happy.

输入描述:

There are multiple test cases. 

The first line of input contains an integer T, indicating the number of test cases. 

For each test case:

The first line contains an integer n

In the next n-1 lines, each line contains two integers u_i and v_i, denoting an edge between monkey u_i and monkey v­_i.

输出描述:

For each test case, output an integer S, denoting the number of triple of monkeys that are happy.

示例1

输入

1
3
1 2
2 3

输出

12

说明

The 12 triples are:
(1,2,1)
(1,2,2)
(1,2,3)
(2,1,1)
(2,1,2)
(2,1,3)
(2,3,1)
(2,3,2)
(2,3,3)
(3,2,1)
(3,2,2)
(3,2,3)

#include<stdio.h>
#include<string.h>
#include<vector>
#include<iostream>
#include<algorithm>
#define N 100005
  
using namespace std;
typedef long long ll;
int dis[N];
  
ll ans;
ll od,ev;
int n;
  
vector<int >ve[N];
  
void dfs(int u,int fa,int dep)
{
    dis[u]=dep;
    int sz=ve[u].size();
    for(int i=0;i<sz;i++){
        int v=ve[u][i];
        if(v==fa) continue;
        dfs(v,u,dep+1);
    }
    return ;
}
  
int main()
{
    int T;
    int u,v;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) ve[i].clear();
        memset(dis,0,sizeof(dis));
        for(int i=1;i<n;i++){
            scanf("%d %d",&u,&v);
            ve[u].push_back(v);
            ve[v].push_back(u);
        }
          
        dfs(1,1,0);
        od=ev=0;
        /*
        for(int i=1;i<=n;i++){
            printf("%d ",dis[i]);
        }
        printf("\n");
        */
        for(int i=1;i<=n;i++){
            if(dis[i]%2==1) od++;
            else ev++;
        }
        ans=0;
          
        if(n%2==1){
            ans=od*ev;
            ans*=2*n;
        }
        else{
            ans+=od*od;
            ans+=ev*ev;
            ans*=n;
             
        }
        printf("%lld\n",ans);
    }
    return 0;
}