P6057 [加油武汉]七步洗手法

Link

Solution:

ans=sum of total triangles - sum of triangles with different colors

Define the angle with different edge as D-angle. Consider triangles with different colors, each such triangle has two D-angles, and these angels don't belong to other triangles.

So compute the numbers of D-angle at each node, then divide by 2, we can get the number of triangles with different colors.

#include <bits/stdc++.h>
# define LL long
using namespace std;

int n,m;
int deg[100005];

int main(){
    scanf("%d %d", &n, &m);
    for(int i=1;i<=m;++i){
        int u,v;
        scanf("%d %d",&u,&v);
        deg[u]++;
        deg[v]++;
    }
    LL res=0;
    for(int i=1;i<=n;++i){
        res+=(LL)deg[i]*(n-1-deg[i]);
    }
    printf("%lld", (LL)n*(n-1)*(n-2)/6-(res>>1));
    return 0;
}