#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
const int N=1e5+10;
int n,m,k;
LL f[N],g[N];
vector<int>a[N],b[N];
struct Node{
int x,y;
Node(){}
Node(int x,int y):x(x),y(y){}
};
Node edge[N];
int main()
{
int Case;scanf("%d",&Case);
while (Case--){
scanf("%d%d%d",&n,&m,&k);
for (int i=1;i<=n;i++)a[i].clear();
for (int i=1;i<=m;i++)b[i].clear();
for (int i=1;i<=k;i++){
int x,y;scanf("%d%d",&x,&y);
a[x].push_back(y);
b[y].push_back(x);
edge[i]=Node(x,y);
}
//f,g
for (int i=1;i<=n;i++){
//boy
f[i]=0;
for (int e=0;e<a[i].size();e++){
f[i]+=b[ a[i][e] ].size()-1;
}
}
for (int i=1;i<=m;i++){
//girl
g[i]=0;
for (int e=0;e<b[i].size();e++){
g[i]+=a[ b[i][e] ].size()-1;
}
}
LL ans=0;
for (int i=1;i<=k;i++){
int x=edge[i].x,y=edge[i].y;
ans+=f[x]-(b[y].size()-1);
ans+=g[y]-(a[x].size()-1);
}
cout<<ans<<endl;
}
return 0;
}
/*
题目: http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=748&pid=1002
题意: n只男羊和m只女羊,在男羊和女羊之间,存在k个朋友关系.以任意一只羊为起点,顺着朋友关系数下去,连续数4只各不相同的羊为1个数羊序列。问有多少种不同的数羊序列。1 <= n, m, k <= 100000
题解: 枚举朋友关系,预先处理出第二只羊有多少个朋友的朋友
*/
BestCoder Round #92-1002 Count the Sheep
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转载自blog.csdn.net/MustImproved/article/details/58628061
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