POJ2002：Squares

标签：poj

Description

Input

Output

Sample Input

4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0

Sample Output

1
6
1

枚举两个相邻的点，求出另外两点，再看这两个点是否存在

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i<=y;i++)
#define mem(a,b) memset(a,b,sizeof(a))
#define w(a) while(a)
const int mod=20007;
int n,next[20007],head[20007],m,ans;
struct node
{
    int x,y;
} a[2222];
int cmp(node a,node b)
{
    if(a.x==b.x)
        return a.y<b.y;
    return a.x<b.x;
}
void insert(int i)
{
    int key=(a[i].x*a[i].x+a[i].y*a[i].y)%mod;
    next[m]=head[key];
    a[m].x=a[i].x;
    a[m].y=a[i].y;
    head[key]=m++;
}

int find(int x,int y)
{
    int key=(x*x+y*y)%mod;
    for(int i=head[key];i!=-1;i=next[i])
    if(a[i].x==x&&a[i].y==y)
    return i;
    return -1;
}

int main()
{
    int i,j;
    w((scanf("%d",&n),n))
    {
        mem(head,-1);
        mem(next,0);
        m=1005;
        ans=0;
        up(i,0,n-1)
        {
            scanf("%d%d",&a[i].x,&a[i].y);
            insert(i);
        }
        sort(a,a+n,cmp);
        up(i,0,n-1)
        {
            up(j,i+1,n-1)
            {
                int x1,y1,x2,y2;
                x1=a[i].x-a[j].y+a[i].y;
                y1=a[i].y+a[j].x-a[i].x;
                if(find(x1,y1)==-1)
                continue;
                x2=a[j].x-a[j].y+a[i].y;
                y2=a[j].y+a[j].x-a[i].x;
                if(find(x2,y2)==-1)
                continue;
                ans++;
            }
        }
        printf("%d\n",ans/2);
    }

    return 0;
}

POJ2002：Squares,布布扣,bubuko.com

POJ2002：Squares

标签：poj

原文地址：http://blog.csdn.net/libin56842/article/details/38638981