2021牛客暑期多校训练营3 J. Counting Triangles（容斥原理）

 
来源：牛客网

题目描述

Goodeat finds an undirected complete graph with n vertices. Each edge of the graph is painted black or white. He wants you to help him find the number of triangles (a, b, c) (a < b < c), such that the edges between (a, b), (b, c), (c, a) have the same color. To avoid the input scale being too large, we use the following code to generate edges in the graph. namespace GenHelper { unsigned z1,z2,z3,z4,b,u; unsigned get() { b=((z1<<6)^z1)>>13; z1=((z1&4294967294U)<<18)^b; b=((z2<<2)^z2)>>27; z2=((z2&4294967288U)<<2)^b; b=((z3<<13)^z3)>>21; z3=((z3&4294967280U)<<7)^b; b=((z4<<3)^z4)>>12; z4=((z4&4294967168U)<<13)^b; return (z1^z2^z3^z4); } bool read() { while (!u) u = get(); bool res = u & 1; u >>= 1; return res; } void srand(int x) { z1=x; z2=(~x)^0x233333333U; z3=x^0x1234598766U; z4=(~x)+51; u = 0; } } using namespace GenHelper; bool edge[8005][8005]; int main() { int n, seed; cin >> n >> seed; srand(seed); for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) edge[j][i] = edge[i][j] = read(); return 0; }
The edge array in the above code stores the color of the edges in the graph. edge[i][j]=1 means that the edge from i to j is black, otherwise it is white (∀0≤i≠j≤n−1∀0≤i\​=j≤n−1).

Ensure that there is an approach that does not depend on the way the data is generated.

输入描述:

The first line contains two integers n(n≤8000),seed(seed≤109)n(n≤8000),seed(seed≤109), denote the number of vertices and the seed of random generator.

输出描述:

Output a line denoting the answer.

示例1

输入

复制

10 114514

输出

复制

35

说明

There're 35 triangles that all three edges have the same color.

考虑容斥，直接用总数减不知足条件的三角形的个数。总数就是\(C_n^3\)，且能够注意到不知足条件的三角形必定是两个边为0一个边为1或者两个边为1一个边为0，即一个三角形中有两个全部关联边异色的顶点和一个全部关联边同色的顶点。而后能够枚举顶点统计关联边异色的顶点个数，除以2就是不知足条件的三角形个数。

#include <bits/stdc++.h>
namespace GenHelper
{
    unsigned z1,z2,z3,z4,b,u;
    unsigned get()
    {
        b=((z1<<6)^z1)>>13;
        z1=((z1&4294967294U)<<18)^b;
        b=((z2<<2)^z2)>>27;
        z2=((z2&4294967288U)<<2)^b;
        b=((z3<<13)^z3)>>21;
        z3=((z3&4294967280U)<<7)^b;
        b=((z4<<3)^z4)>>12;
        z4=((z4&4294967168U)<<13)^b;
        return (z1^z2^z3^z4);
    }
    bool read() {
      while (!u) u = get();
      bool res = u & 1;
      u >>= 1; return res;
    }
    void srand(int x)
    {
        z1=x;
        z2=(~x)^0x233333333U;
        z3=x^0x1234598766U;
        z4=(~x)+51;
      	u = 0;
    }
}
using namespace GenHelper;
using namespace std;
bool edge[8005][8005];
int main() {
    int n, seed;
    cin >> n >> seed;
    srand(seed);
    for (int i = 1; i <= n; i++) {
    	for (int j = i + 1; j <= n; j++) {
        	edge[j][i] = edge[i][j] = read();
        }
    }
    long long cnt = 0, all = 0;
    for(int i = 1; i <= n; i++) {
        long long tmp = 0;
        long long x = 0, y = 0;
        for(int j = 1; j <= n; j++) {
            if(i == j) continue;
            if(edge[i][j]) x++;
            else y++;
        }
        cnt += x * y;
    }
    cout << 1ll * n * (n - 1) * (n - 2) / 6 - cnt / 2;
 	return 0;
}