Question

Bitset是检查点是否形成Square的最有效方法。

读取对（x，y）的N值，使用bitset计算N对形成的平方数（和最大面积）的最佳方法是什么？

输入：

输出：

3 -> Number of squares formed by the pairs x,y
5 -> biggest area

O（N ^ 2）解决方案是最有效的方法吗？

我的代码到目前为止：我认为不正确：

#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;

struct c{

    int x,y;

}A,B,coordenadas[3001],R[4];

struct cc{

    long int c1,c2;

}R1,R2;

int quadrados=0,n=0;
long int area;
long int concat[3001];

int bs(long int search)
{
    int first = 0;
    int last = n - 1;
    int middle = (first+last)/2;

    while( first <= last )
    {
      if ( concat[middle] < search )
         first = middle + 1;
      else if ( concat[middle] == search )
      {
         return 1;
         break;
      }
      else
         last = middle - 1;

      middle = (first + last)/2;
    }
    if ( first > last )
      return 0;
}

void calculo()
{
    int x=0,y=0,a=0,b=0,limpa=0;


    if(A.y==B.y)
    {
        x=abs(A.x-B.x);
        area=x*x;
        R1.c1=(B.y + x) * 100000 + (A.x);
        R1.c2=(B.y - x) * 100000 + (A.x);
        //R2.c1=(B.y + x) * 100000 + (B.x);
        //R2.c2=(B.y - x) * 100000 + (B.x);
        if((int)bs(R1.c1)==1 && (int)bs(R1.c2)==1)
        quadrados+=1;
        else
        area=0;
    }
    else
    if(A.x==B.x)
    {
        x=abs(B.y-A.y);
        area=x*x;
        R1.c1=(B.y) * 100000 + (A.x + x);
        R1.c2=(B.y) * 100000 + (A.x - x);
        //R2.c1=(A.y) * 100000 + (B.x + x);
        //R2.c2=(A.y) * 100000 + (B.x - x);
        if((int)bs(R1.c1)==1 && (int)bs(R1.c2)==1)
        quadrados+=1;
        else
        area=0;

    }
    else
    if(A.y>B.y)
    {
        x=abs(B.x-A.x);
        y=abs(B.y-A.y);
        area=sqrt(x*x+y*y)*sqrt(x*x+y*y);
        R1.c1=(A.y + x) * 100000 + (A.x + y);
        R1.c2=(B.y + x) * 100000 + (B.x + y);
        //R2.c1=(A.y - x) * 100000 + (A.x - y);
        //R2.c2=(B.y - x) * 100000 + (B.x - y);
        if((int)bs(R1.c1)==1 && (int)bs(R1.c2)==1)
        quadrados+=1;
        else
        area=0;
    }
    else
    if(A.y<B.y)
    {
        x=abs(B.y-A.y);
        y=abs(B.x-A.x);
        area=sqrt(x*x+y*y)*sqrt(x*x+y*y);
        R1.c1=(B.y + y) * 100000 + (B.x - x);
        R1.c2=(A.y + y) * 100000 + (A.x - x);
        //R2.c1=(B.y - y) * 100000 + (B.x + x);
        //R2.c2=(A.y - y) * 100000 + (A.x + x);
        if((int)bs(R1.c1)==1 && (int)bs(R1.c2)==1)
        quadrados+=1;
        else
        area=0;

    }

//  printf("Area=%ld e R0-> (%d,%d) e R1 (%d,%d) R2 (%d,%d) R3 (%d,%d)\n",area,R[0].x,R[0].y,R[1].x,R[1].y,R[2].x,R[2].y,R[3].x,R[3].y);
}

int main() {

    int M=1,mArea=0;
    cin >> n;
    for(int i=0;i<n;i++)
    {cin >> coordenadas[i].x >> coordenadas[i].y;concat[i]=coordenadas[i].y * 100000 + coordenadas[i].x;}
    for(int i=0;i<n;i++)
    {
        A.x=coordenadas[i].x;A.y=coordenadas[i].y;
        for(int ii=M;ii<n;ii++)
        {
            B.x=coordenadas[ii].x;B.y=coordenadas[ii].x;
            calculo();
            if(mArea<area)
            mArea=area;
        }
        M+=1;
    }

    if(quadrados==0)
    cout << quadrados << endl;
    else
    cout << quadrados << endl << mArea << endl;


    return 0;
}

如何使用Bitset C ++检查点是否形成正方形

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