17暑假多校联赛2.11 HDU 6055 Regular polygon

Regular polygon

Time Limit: 3000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
 

Problem Description

On a two-dimensional plane, give you n integer points. Your task is to figure out how many different regular polygon these points can make.
 

Input

The input file consists of several test cases. Each case the first line is a numbers N (N <= 500). The next N lines ,each line contain two number Xi and Yi(-100 <= xi,yi <= 100), means the points’ position.(the data assures no two points share the same position.)
 

Output

For each case, output a number means how many different regular polygon these points can make.
 

Sample Input

40 00 11 01 160 00 11 01 12 02 1

Sample Output

12

 
题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=6055
 

分析

题意：给出n个点，判断这些点能够构成多少个正多边形
形成正多边形需要每条边都相等，因为没点坐标都为整数，所以能构成的正多边形只有正方形，可以先将各个点排序可以按照先横坐标再纵坐标，也可以按照先纵坐标再横坐标，然后先选中两个点再往这两个点的上方或者右上方确定能组成正方形的点的坐标，然后判断这两个坐标是否在已给的坐标中
 

代码

#include <iostream>#include <stdio.h>#include <math.h>#include <string.h>#include <algorithm>#define pf(a) a*ausing namespace std;int N;struct point{    int x,y;} p[510];///存放给的点的坐标bool cmp(struct point a,struct point b){    if(a.y==b.y)///按照在坐标系中的位置先下再左的顺序排序    {        return a.x<b.x;    }    else return a.y<b.y;}bool findp(int x,int y)///查找点(x,y)是否在给定的点中{    for(int i=0; i<N; i++)    {        if(y==p[i].y&&x==p[i].x)        {            return true;        }    }    return false;}int main(){    while(~scanf("%d",&N))    {        int sum=0;        for(int i=0; i<N; i++)        {            scanf("%d %d",&p[i].x,&p[i].y);        }        sort(p,p+N,cmp);///将给的点排序        for(int i=0; i<N-1; i++)        {            for(int j=i+1; j<N; j++)///让所有点都相互连接来作为正方形的一条边            {                if(p[i].y==p[j].y||(p[j].y>p[i].y&&p[j].x<p[i].x))///排序之后点p[j]的y值总是大于等于p[i]的y，因为要避免重复查点，所以只查询上方和右上方的可以构成正方形的点，所以p[j]的y大于p[i]的y时，它的x要小于p[i]的x，然后查找满足构成正方形的点                {                    if(p[i].y==p[j].y)                    {                        if((findp(p[i].x,(p[i].y+abs(p[i].x-p[j].x))))&&findp(p[j].x,(p[j].y+abs(p[i].x-p[j].x))))                            sum++;                    }                    else                    {                        if((findp((p[i].x+abs(p[i].y-p[j].y)),(p[i].y+abs(p[i].x-p[j].x))))&&(findp((p[j].x+abs(p[i].y-p[j].y)),(p[j].y+abs(p[i].x-p[j].x)))))                            sum++;                    }                }            }        }        printf("%d\n",sum);    }}