Counting Triangles（逆向思维）2021牛客暑期多校训练营3

J. Counting Triangles

题意：

​ (a, b, c) (a < b < c)，(a, b), (b, c), (c, a) 三边同时为true or false 计数++

复盘：

​ a -> b -> c -> a , 形成了一个三角形，首先肯定能想到 n 三次暴力枚举，必然超时，想半天也没什么思路，经队友提醒，首先是无向图（一直拿有向图去写），然后三角形的三边情况就四种，全1，全0，两1，两0。三角形总个数为Cn取3，去算不计数的个数，不计数的个数为：对于每一个点，分别计数其到剩下的点，边权为0和1的个数，两者相乘即为不计数个数，遍历一遍求sum再/2(i到j，j到i 重复了)

#include <bits/stdc++.h>
#define int 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;
using namespace std;
bool edge[8005][8005];
signed 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();
    int sum=0;
    for(int i=0;i<n;i++)
    {
        int se=0, sl=0;
        for(int j=0;j<n;j++)
        {
            if(i == j) continue;
            if(edge[i][j]) se++;
            else sl++;
        }
        sum += se*sl;
    }
    cout<<n*(n-1)*(n-2)/6-sum/2<<endl;

    return 0;
}

标签：int,++,多校,牛客,2021,z4,z1,z2,z3
来源： https://blog.csdn.net/qq_51354600/article/details/119064140