POJ 2002 Squares(简单哈希表的运用)

版权声明：听说这里让写版权声明～～～ https://blog.csdn.net/m0_37691414/article/details/82153518

题意：给定n个相异整数点，求能构成多少个正方形。

其实这题挺简单的，经过思考可以发现，我们可以通过枚举两个点（或者说一条边），然后求出可能构成正方形的另外两个点，然后判断是不是在给你的点集中就行了。求点问题是一个数学问题，判断要用到hash表。用开散列法(拉链法).

#include<stdio.h>
#include<string.h>
#include <iostream>
using namespace std;
#define MAXHASH 1000003
#define NN 1004
typedef struct node{
    int x, y;
    struct node *nxt;
}NODE;

NODE po[NN];
NODE *head[MAXHASH];
int n;

int hash(NODE p){
    int h = p.x * 20000 + p.y;
    h %= MAXHASH;
    if(h < 0) h += MAXHASH;
    return h;
}
void Init(){
    int i, h;
    memset(head, 0, sizeof(head));
    for (i = 1; i <= n; i++){
        h = hash(po[i]);
        po[i].nxt = head[h];
        head[h] = po + i;
    }
}

int find(NODE p1){
    int h = hash(p1);
    for (NODE *p = head[h]; p; p = p->nxt){
        if(p->x == p1.x && p->y == p1.y){
            return 1;
        }
    }
    return 0;
}
void Solve(){

    NODE p1, p2, p3, p4;
    int i, j;
    int cnt = 0;
    for (i = 1; i <= n; i++){
        for (j = 1; j <= n; j++){
            if(i == j) continue;
            //公式：x3=x1+(y1-y2);y3=y1-(x1-x2); x4=x2+(y1-y2);y4=y2-(x1-x2)(在直线AB上方)
            //    或x3=x1-(y1-y2);y3=y1+(x1-x2); x4=x2-(y1-y2);y4=y2+(x1-x2)(在直线AB下方)
            p1 = po[i];
            p2 = po[j];
            p3.x = p1.x + (p1.y - p2.y);
            p3.y = p1.y - (p1.x - p2.x);
            p4.x = p2.x + (p1.y - p2.y);
            p4.y = p2.y - (p1.x - p2.x);

            if(find(p3) && find(p4)) cnt++;
        }
    }
    printf("%d\n", cnt / 4);
}
int main() {
    int i;
    while(scanf("%d", &n)!= EOF){
        if(n == 0) break;
        for (i = 1; i <= n; i++){
            scanf("%d%d", &po[i].x, &po[i].y);
        }
        Init();
        Solve();
    }
    return 0;
}