原題連結

2021牛客暑期多校訓練營3

J.Counting Triangles

時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld

題目描述

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 \((\forall 0 \le i \neq j \le n-1).\)
Ensure that there is an approach that does not depend on the way the data is generated.

輸入描述:

The first line contains two integers \(n(n \le 8000), seed (seed \le 10^9)\)

輸出描述:

Output a line denoting the answer.

輸入

10 114514

輸出

35

說明

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

解題思路

容斥原理

原問題不好直接求解，我們不妨考慮其反面，即：異色三角形
我們只需要找出這樣的點的數量：其兩邊異色，找出所有這樣的點的數量 \(cnt\) 後，\(cnt \div 2\) 即異色三角形的個數，用總個數 \(C_n^3\) 減去即可~

\(\color{red} {為什麼 cnt \div 2 即異色三角形的個數？}\)
我們求得的 \(cnt\) 是異色角的個數，而一個異色三角形由兩個異色角和一個同色角構成，每個點列舉的時候都會重複列舉一次
注意：可能會超int

• 時間複雜度:\(O(n^2)\)

程式碼

#include<bits/stdc++.h>
using namespace std;
using LL=long long;
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();
    LL diff=0;
    for(int i=0;i<n;i++)
    {
        LL s[2]={0,0};
        for(int j=0;j<n;j++)
        {
            if(i==j)continue;
            s[edge[i][j]]++;
        }
        diff+=s[0]*s[1]/2;
    }
    cout<<1ll*n*(n-1)*(n-2)/6-diff;
 	return 0;
}