poj 3275（邻接表解法，first，next，v代表以k为起点的终点集，first2，next2，v2代表以k为终点的起点集）

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn = 1005;
const int N = 1000005;
int data[maxn][maxn];
int next[N],v[N],first[N],next2[N],v2[N],first2[N];
int n,m,total,total2;
void add(int a,int b){
    v[++total] = b;
    next[total] = first[a];
    first[a] = total;
}
void add2(int a,int b){
    v2[++total2] = b;
    next2[total2] = first2[a];
    first2[a] = total2;
}
int main(){
    int i,j,a,b,ans;
    scanf("%d%d",&n,&m);
    memset(data,0,sizeof data);
    for(i=1;i<=m;i++){
        scanf("%d%d",&a,&b);
        data[a][b] = 1;
        add(a,b);
        add2(b,a);
    }
    ans = m;
    for(int k=1;k<=n;k++){
        for(int i=first[k];i>0;i=next[i]){
            for(int j=first2[k];j>0;j=next2[j]){
                if(data[v2[j]][v[i]]==0){
                    data[v2[j]][v[i]] = 1;
                    ans++;
                    add(v2[j],v[i]);
                    add2(v[i],v2[j]);
                }
            }
        }
    }
    cout<<n*(n-1)/2-ans<<endl;
    return 0;
}