POJ_2002Squares解题报告
Squares
| Time Limit: 3500MS | Memory Limit: 65536K | |
| Total Submissions: 11181 | Accepted: 4064 |
Description
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.
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.
Input
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.
Output
For each test case, print on a line the number of squares one can form from the given stars.
Sample Input
4 1 0 0 1 1 1 0 0 9 0 0 1 0 2 0 0 2 1 2 2 2 0 1 1 1 2 1 4 -2 5 3 7 0 0 5 2 0
Sample Output
1 6 1
题目连接：http://poj.org/problem?id=2002
算法类型：Hash
解题思路：先把每个点所对应的Hash值储存在对应的容器里，在依次枚举两个点，算出其所对应的另外四个点，分别在其对应的容器中依次查找，如果两个点都找到，总数加一，最后把总数除以四。
算法实现：
#include
#include
#include
#include
#include
using namespace std;
#define MAX 9997
struct Squares
{
int x;
int y;
}squ[100002];
int Hash(Squares squ) //hash 函数；
{
int s;
s=squ.x*squ.x+squ.y*squ.y+squ.y;
return s%MAX;
}
int main()
{
int N;
while(scanf("%d",&N)!=EOF)
{
if(N==0)
break;
memset(squ,0,100002);
vector v[MAX]; //系统容器；
memset(v,0,MAX);
int sum=0;
int i,J,JJ,P,PP;
int hash,hash1,hash2,hash3,hash4;
int X1,Y1,X2,Y2,X3,Y3,X4,Y4;
for(i=0;i