HDU 6055 Regular polygon（计算几何+思维）——2017 Multi-University Training Contest

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Regular polygon

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1187    Accepted Submission(s): 461

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

4 
0 0 
0 1 
1 0 
1 1 
6 
0 0 
0 1 
1 0 
1 1 
2 0 
2 1

 

Sample Output

1 
2

题目大意：

给了 n(n≤500) 个整数点，求能构成多少个正多边形

解题思路：

首先因为是整数点，所以只能构成正四边形也就是正方形，首先想过暴力，复杂度 O(n4)， 爆炸。 然后想的是 如何优化。因为是正方形，我们可以直接枚举两个点，然后判断剩余的两个点是否存在就可以了。
首先我们按照 x 坐标从小到大进行排序，然后计算另外两个点的坐标，然后判断，需要注意的是我们多算了一遍，要除以2
有一个坑点：就是将负数点转化为正数的时候，经过计算：要加上>300 的数。

代码：

#include <bits/stdc++.h>using namespace std;const int MAXN = 605;int mp[MAXN][MAXN];struct Point{    int x, y;}p[MAXN];int cmp(Point A, Point B){    if(A.x == B.x) return A.y < B.y;    return A.x < B.x;}int main(){    ///freopen("in.txt","r", stdin);    int n;    while(~scanf("%d", &n)){        memset(mp, 0, sizeof(mp));        for(int i=0; i<n; i++) {            int x, y;            scanf("%d%d", &x, &y);            p[i].x = (x + 300);            p[i].y = (y + 300);            mp[p[i].x][p[i].y] = 1;        }        sort(p, p+n, cmp);        int sum = 0;        for(int i=0; i<n; i++){            for(int j=i+1; j<n; j++){                int dx = p[i].x - p[j].x;                int dy = p[j].y - p[i].y;                if(mp[p[i].x+dy][p[i].y+dx] && mp[p[j].x+dy][p[j].y+dx]) sum++;            }        }        printf("%d\n",sum/2);    }    return 0;}