2017 杭电多校联赛第二场 1011 Regular polygon（多个点求正方形个数）POJ 2002

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

40 00 11 01 160 00 11 01 12 02 1

 

Sample Output

12
跟POJ 2002一模一样。。。
则p1 p2 p3 p4组成一个正方形，设p1 = (x1,y1), p2 = (x2, y2)，根据向量的旋转公式可以求出p3, p4的坐标为
p3 = (y1 - y2 + x1, x2 - x1 + y1)
p4 = (y1 - y2 + x2, x2 - x1 + y2)
#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int N = 1010;const int H = 10007;int ptx[N], pty[N];struct Node{    int x;    int y;    int next;};Node node[N];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;}