BestCoder Round #92-1002 Count the Sheep

#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;
}
/*
題目： http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=748&pid=1002
題意： n只男羊和m只女羊，在男羊和女羊之間，存在k個朋友關係.以任意一隻羊爲起點，順着朋友關係數下去，連續數4只各不相同的羊爲1個數羊序列。問有多少種不同的數羊序列。1 <= n, m, k <= 100000

題解： 枚舉朋友關係，預先處理出第二隻羊有多少個朋友的朋友
*/