POJ2002_Squares (哈希表)

本文出自：blog.csdn.net/svitter

题意

A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property.
So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates.

一共可以在给出的n个点中找出多少个正方形？

输入输出分析

The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow.
Each of the next n lines specify the x and y coordinates (two integers) of each point.
You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.

有多组测试数据，每一个测试数据n开始，n <1000说明最多有1000个点。坐标系的横纵坐标不会超过20000, 以n=0结束

算法数据结构分析

本题目用哈希表+简单的分析做～

给出4个点，如果四重循环N**4必然超时，那么枚举两个点来计算出对应的两个点，计算对应点是否存在



计算方法见

getPoint()

函数

计算定应点是否存在的最快速度就是使用hash来查找。把（x+y）mod m，即取余法。最多有1000个点，所以哈希表1000个slot足够。
保险起见，开1010，M取质数1009。

使用拉链法解决冲突，开一个next数组，来保存下标i对应到hash表上，因为最多1000个节点，最糟糕的情况下需要999个卫星表， 所以next数组开·1010即可。

AC代码：

//author: svtter
//

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <map>
#include <algorithm>
#include <queue>
#include <cmath>

#define INF 0xffffff
#define lln long long

#ifdef ONLINE_JUDGE
#define FOI(file) 0
#define FOW(file) 0
#else
#define FOI(file) freopen(file,"r",stdin);
#define FOW(file) freopen(file,"w",stdout);
#endif

using namespace std;

#define MAXN 1010
#define MOD 1009
struct node
{
int x, y;
bool operator == (const node &a)const
{
return (x == a.x && y == a.y);
}
bool operator < (const node &a)const
{
if(x!=a.x)
return x <a.x;
return y < a.y;
}
};
node point[MAXN];
int hash[MAXN+1], next[MAXN];

int n;

node c, d;
void getPoint(node &a, node &b)
{
c.x = b.x + (b.y - a.y);
c.y = b.y - (b.x - a.x);
d.x = a.x + (b.y - a.y);
d.y = a.y - (b.x -a.x);
}

int getHash(node &a)
{
return abs(a.x+ a.y) % MOD;
}

void insertHash(int i)
{
//头插入，拉链法
int key = getHash(point[i]);
next[i] = hash[key];
hash[key] = i;
}

bool search(node &t)
{
int key = getHash(t);
int i = hash[key];
while(i != -1)
{
if(point[i] == t)
return true;
i = next[i];
}
return false;
}

int main()
{
FOI("input");
//FOW(output);

int i, j;
int ans;
//write your programme here
while(~scanf("%d", &n))
{
if(n == 0)
break;
memset(hash , -1 ,sizeof(hash));
memset(next , -1, sizeof(next));
for(i = 0; i < n; i++)
scanf("%d%d", &point[i].x, &point[i].y);
sort(point ,point+n);
//puts("input over.");

for(i = 0; i < n; i++)
insertHash(i);
//puts("build hash over.");

ans = 0;
for(i = 0; i < n; i++)
{
for(j = i+1; j < n; j++)
{
getPoint(point[i], point[j]);
if(search(c) && search(d))
ans++;
}
}

printf("%d\n", ans/2);
}

return 0;
}