2017杭电多校联赛第二场-Regular polygon （hdu6055）判断点集能构成多少个正方形

Regular polygon

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

Total Submission(s): 0    Accepted Submission(s): 0


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：给你n个点，询问可以构成多少个正方形
解题思路：由题，有n个点，询问可以构成正方形的个数，其实我们只需要判断有没有点能构成正方形，因为正方形的形状特殊性，所以我们知道了正方形的两个点就可以得到其余的两个点，所以我们只需要判断另外两个点在不在我们的点集中即可。
ac代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const int N = 1010;
const int H = 10007;
int ptx
, pty
;

struct Node
{
int x;
int y;
int next;
};
Node node
;
int cur;
int n;
long ans;
int hashTable[H];

void initHash()
{
for (int i = 0; i < H; ++i) hashTable[i] = -1;
cur = 0;
ans = 0;
}

void insertHash(int x, int y)
{
int h = (x * x + y * y) % H;
node[cur].x = x;
node[cur].y = y;
node[cur].next = hashTable[h];
hashTable[h] = cur;
++cur;
}

bool searchHash(int x, int y)
{
int h = (x * x + y * y) % H;
int next;
next = hashTable[h];
while (next != -1)
{
if (x == node[next].x && y == node[next].y) return true;
next = node[next].next;
}
return false;
}

int main()
{
while (scanf("%d", &n) != EOF && n)
{
initHash();
for (int i = 0; i < n; ++i)
{
scanf("%d%d", &ptx[i], &pty[i]);
insertHash(ptx[i], pty[i]);
}
for (int i = 0; i < n; ++i)
{
for (int j = i + 1; j < n; ++j)
{
int x1 = ptx[i] - (pty[i] - pty[j]);
int y1 = pty[i] + (ptx[i] - ptx[j]);
int x2 = ptx[j] - (pty[i] - pty[j]);
int y2 = pty[j] + (ptx[i] - ptx[j]);
if (searchHash(x1, y1) && searchHash(x2, y2)) ++ans;
}
}
for (int i = 0; i < n; ++i)
{
for (int j = i + 1; j < n; ++j)
{
int x1 = ptx[i] + (pty[i] - pty[j]);
int y1 = pty[i] - (ptx[i] - ptx[j]);
int x2 = ptx[j] + (pty[i] - pty[j]);
int y2 = pty[j] - (ptx[i] - ptx[j]);
if (searchHash(x1, y1) && searchHash(x2, y2)) ++ans;
}
}
ans >>= 2;
printf("%ld\n", ans);
}
return 0;
}

题目链接：点击打开链接http://acm.hdu.edu.cn/showproblem.php?pid=6055