hdu6055(求正方形个数)

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

题意: 给出 n 组坐标 x, y, 输出其中的正多边形个数 . 其中 x, y 均为整数.

思路: x, y 为整数, 所以只存在正方形, 不会有其他正多边形 . 那么只需要枚举正方形的对角线即可 .

代码:

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define ll long long
using namespace std;

const int MAXN = 1e3 + 10;
const int H = 1e4 + 7;

struct node{
    int x, y, next;
}gel[MAXN];

ll ans;
int head[H];
int indx, n;
int ptx[MAXN], pty[MAXN];

void init(void){
    memset(head, -1, sizeof(head));
    indx = 0;
    ans = 0;
}

void add(int x, int y){
    int h = (x * x + y * y) % H;
    gel[indx].x = x;
    gel[indx].y = y;
    gel[indx].next = head[h];
    head[h] = indx++;
}

bool find(int x, int y){
    int h = (x * x + y * y) % H;
    for(int i = head[h]; i != -1; i = gel[i].next){
        if(x == gel[i].x && y == gel[i].y) return true;
    }
    return false;
}

int main(void){
    while(~scanf("%d", &n)){
        init();
        for(int i = 0; i < n ; i++){
            scanf("%d%d", &ptx[i], &pty[i]);
            add(ptx[i], pty[i]);
        }
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if(i == j) continue;
                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(find(x1, y1) && find(x2, y2)) ans++;
            }
        }
        printf("%lld
", ans >> 2);
    }
    return 0;
}