一起数羊 hhhhhhhhB - Count the Sheep

Altough Skipping the class is happy, the new term still can drive luras anxious which is of course because of the tests! Luras became worried as she wanted to skip the class, as well as to attend the BestCoder and also to prepare for tests at the same time. 

However, As the result of preparing for tests, luras had no time to practice programing. She didn't want to lose her rating after attending BC. In the end, she found BCround92's writer snowy_smile for help, asking him to leak her something. 

Snowy_smile wanted to help while not leaking the problems. He told luras, the best thing to do is to take a good rest according to the following instructions first. 

"Imagine you are on the endless grassland where there are a group of sheep. And n sheep of them are silent boy-sheep while m sheep are crying girl-sheep. And there are k friend-relationships between the boy-sheep and girl-sheep.Now You can start from any sheep, keep counting along the friend relationship. If you can count 4 different sheep, you will exceed 99% sheep-counters and fall asleep." 

Hearing of the strange instructions, luras got very shocked. Still, she kept counting. Sure enough, she fell asleep after counting 4 different sheep immediately. And, she overslept and missed the BestCoder in the next day. At a result, she made it that not losing her rating in the BCround92!!! 

However, you don't have the same good luck as her. Since you have seen the 2nd problem, you are possible to have submitted the 1st problem and you can't go back. 

So, you have got into an awkward position. If you don't AC this problem, your rating might fall down. 

You question is here, please, can you tell that how many different 4-sheep-counting way luras might have before her sleep? 

In another word, you need to print the number of the "A-B-C-D" sequence, where A-B, B-C, C-D are friends and A,B,C,D are different.

InputThe first line is an integer T which indicates the case number. 

and as for each case, there are 3 integers in the first line which indicate boy-sheep-number, girl-sheep-number and friend-realationship-number respectively. 

Then there are k lines with 2 integers x and y in each line, which means the x-th boy-sheep and the y-th girl-sheep are friends. 

It is guaranteed that—— 

There will not be multiple same relationships. 

1 <= T <= 1000 

for 30% cases, 1 <= n, m, k <= 100 

for 99% cases, 1 <= n, m, k <= 1000 

for 100% cases, 1 <= n, m, k <= 100000OutputAs for each case, you need to output a single line. 

there should be 1 integer in the line which represents the number of the counting way of 4-sheep-sequence before luras's sleep.Sample Input

3
2 2 4
1 1
1 2
2 1
2 2
3 1 3
1 1
2 1
3 1
3 3 3
1 1
2 1

2 2

本来没思路 看了一篇大佬的博客 顿悟啊 大佬就是大佬啊 hhhhhhhhh

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <map>
using namespace std;
///通过一组有联系的公母羊 再分别找一个公母羊  便是两条路径
int a[100005];
int b[100005];
struct ss
{
    int x, y;
}s[100005];///记录有联系的公母羊
int main()
{
    int t;
    scanf("%d", &t );
    while( t-- )
    {
        int i , x ,y;
        memset( a, 0, sizeof(a ));
        memset( b, 0, sizeof( b ));
        int n, m, k;
        scanf("%d%d%d", &n, &m, &k );
        for( i =0 ;i<k; i++)
        {
            scanf("%d%d", &x, &y);
            s[i].x = x;
            s[i].y = y;
            a[x]++;///x公羊的有联系母羊个数
            b[y]++;///y母羊的有联系的公羊个数
        }
        long long int sum= 0;
        for( i = 0; i<k; i++)
        {
            sum = sum + ( a[s[i].x]-1 )*( b[s[i].y]-1);///和这只公羊有联系的母羊个数-1 （删除现在这只） 乘以和这只母羊
            ///有联系的公羊个数 再乘以2即可 （ 来回两种路径）
        }
        printf("%lld\n", sum*2);
    }
}