Counting Triangles

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\forall 0 \le i \neq j \le 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 \le 8000), seed (seed \le 10^9)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. 

思路：

注意到⼀个神奇的性质：每个三⻆形要么同⾊，要么有两边同⾊另⼀边异⾊。对于后者，三⻆形有恰有两个异⾊ ⻆，⽽前者没有异⾊⻆。 因此异⾊⻆数 /2 即为不符合条件的三⻆个数。⽤总数减去即可。 所以我们可以根据边的颜色先把顶点颜色记录，后面枚举计算异色个数。 而凸多边形里的三角形个数为组合问题：C(n,3)=n*(n-1)(n-2)/6  代码如下： #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
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;
ll n,num,ans;
int seed;
ll s[8005][2];
int main(){
    cin>>n>>seed;
    srand(seed);
    for(int i=0;i<n;i++){
        for(int j=i+1;j<n;j++){
            int x=read();//1:black, 0:white
            s[i][x]++;//记录顶点颜色 
            s[j][x]++;
        }
    }
    for(int i=0;i<n;i++){
        num+=s[i][0]*s[i][1];//异色角数量 
    }
    ans=n*(n-1)*(n-2)/6-num/2;//总数：n*(n-1)*(n-2)/6-异色角数量/2 
    cout<<ans<<endl;
    return 0;
}

标签：int,seed,z2,z4,Counting,z1,Triangles,z3
来源： https://blog.csdn.net/srh20/article/details/119063939