BOJ 1575 Triangles

Triangles

Accept:12     Submit:28

Time Limit:1000MS     Memory Limit:65536KB

Description

There are points given in a space. There are no three points, such that they lie on the same straight line. Each pair of points is connected by a segment coloured red or black. Each triangle, whose sides have the same colour is called a monochromatic triangle. We are given a list of all red segments and we want to find the number of all monochromatic triangles.

 

InputFormat

The first line is an integer T(1≤T≤20) indicating the case number.

For each case,the first line is N(1≤N≤1000) indicating the number of points and M(1≤M≤1000000) indicating the number of red edges.

In each of the following M lines there are two integers x and y separated by a single space(1≤x,y≤N). They are numbers of vertices which are end points of a red segment.

 

OutputFormat

For each case output the number of monochromatic triangles(which all 3 edges are same color).

 

SampleInput

1

6 9

1 2

2 3

2 5

1 4

1 6

3 4

4 5

5 6

3 6

SampleOutput

2

第三场新生排位赛的C题，比赛时看题面就吓到了，但后来听学长讲过后，发现经过转换后是一道很简单的数学题。

给出n个点构成的一个完全图，图中的边有两种颜色，红色和黑色，求同色边构成的三角形有多少个。

首先我们知道图中三角形总数为C(n,3)

对于每一个点i，我们可以统计出由它连出去的红色边数red[i]和黑色变数n-1-red[i]，这样该点连出去的不同色三角形就是red[i]*(n-1-red[i])

将所有点连出去的不同色三角形数目1求和再除以2（因为每条边被计算了两次）就得到图中不同色三角形的总数。

三角形总数减去不同色三角形数就是同色三角形数

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

long red[1001];

int main()
{
    int t;
    long m,n;

    scanf("%d",&t);

    while(t--)
    {
        int a,b;
        long ans,total,dif=0;

        memset(red,0,sizeof(red));

        scanf("%ld %ld",&n,&m);
        total=n*(n-1)*(n-2)/6;

        for(int i=1;i<=m;i++)
        {
            scanf("%d %d",&a,&b);
            red[a]++;
            red[b]++;
        }

        for(int i=1;i<=n;i++)
            dif+=red[i]*(n-1-red[i]);
        ans=total-dif/2;

        printf("%ld\n",ans);
    }

    return 0;
}