链接：http://poj.org/problem?id=2002
Squares
| Time Limit: 3500MS | Memory Limit: 65536K | |
| Total Submissions: 21720 | Accepted: 8321 |
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.
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
先留上几个坑：
①据说有二分做法，占坑待更；
②一开始用的向量,T了。。。回头把vector换成数组，然后优化一下常数，占坑待更
题目大意：
平面内有n(1<=n<=1000)个点，给定点的坐标Xi,Yi,求一共可以组成多少个正方形
分析：根据几何关系，由其中两点a,b确定另外两点m,n,看m,n是否存在，确定两点的时间复杂度是O(n2),查找的时间复杂度是O(logN),总体是O(2n2logn)数据范围在1000内，可以接受
一开始用的向量匹配，确定n2条向量，然后确定令一条向量的起点和大小，方向（平行），进行向量匹配，然后T了(坑①)（这里还有个（错误的）思路,两个向量垂直，不过这样只能确定三个点，就没再想了）然后就根据两个点匹配另外两个点，900+MS,A掉了(看来常数写的不行)
上代码：
1 #include<iostream>
2 #include<cstdio>
3 #include<queue>
4 #include<cstdio>
5 #include<cstring>
6 #include<cstdlib>
7 #define mem(a,b) memset(a,b,sizeof(a))
8 using namespace std;
9
10 const int mod=100007;
11 const int maxn=120009;
12 struct node
13 {
14 int x,y,next;
15 }edge[maxn];
16 int cnt;
17 int ans,head[maxn];
18 int xi[1111],yi[1111];
19 inline int read()
20 {
21 char ch = getchar(); int x = 0, f = 1;
22 while(ch < ‘0‘ || ch > ‘9‘) {if(ch == ‘-‘) f = -1; ch = getchar();}
23 while(‘0‘ <= ch && ch <= ‘9‘) {x = x * 10 + ch - ‘0‘; ch = getchar();}
24 return x * f;
25 }
26 void out(int a) //代替printf，速度更快，俗称输出挂
27 {
28 if(a > 9)
29 {
30 out(a/10);
31 }
32 putchar(a%10 + ‘0‘);
33 }
34 void ins(int x,int y,int h)
35 {
36 h%=mod;
37 edge[cnt].x=x;
38 edge[cnt].y=y;
39 edge[cnt].next=head[h];
40 head[h]=cnt++;
41 }
42 inline bool cmp(node a,node b)
43 {
44 if(a.x==b.x&&a.y==b.y) return 1;
45 return 0;
46 }
47 bool search(node a,int h)
48 {
49 h%=mod;
50 //printf("search:\n");
51 for(int i=head[h];i!=-1;i=edge[i].next)
52 {
53 //"%d %d %d %d %d %d %d %d %d %d %d %d\n",xi[i],yi[i],xi[j],yi[j],a.x,a.y,b.x,b.y,c.x,c.y,d.x,d.y);
54 if(cmp(edge[i],a)) return 1;
55 }
56 return 0;
57 }
58 void init()
59 {
60 mem(edge,0);
61 mem(head,-1);
62 mem(xi,0);
63 mem(yi,0);
64 cnt=0;
65 ans=0;
66 }
67 int main()
68 {
69 int n;
70 while(scanf("%d",&n)&&n)
71 {
72 init();
73 for(int i=1;i<=n;i++)
74 {
75 xi[i]=read();yi[i]=read();
76 ins(xi[i],yi[i],abs(xi[i]*1000+yi[i]));
77 }
78 for(int i=1;i<=n;i++)
79 {
80 for(int j=i+1;j<=n;j++)
81 {
82 int dty=yi[j]-yi[i],dtx=xi[j]-xi[i];
83 node a,b,c,d;
84 a.x=xi[i]+dty;a.y=yi[i]-dtx;
85 b.x=xi[j]+dty;b.y=yi[j]-dtx;
86 c.x=xi[i]-dty;c.y=yi[i]+dtx;
87 d.x=xi[j]-dty;d.y=yi[j]+dtx;
88 int ha=abs(a.x*1000+a.y),hb=abs(b.x*1000+b.y),
89 hc=abs(c.x*1000+c.y),hd=abs(d.x*1000+d.y);
90 //printf("%d %d %d %d %d %d %d %d %d %d %d %d\n",xi[i],yi[i],xi[j],yi[j],a.x,a.y,b.x,b.y,c.x,c.y,d.x,d.y);
91 if(search(a,ha)&&search(b,hb)) {ans++;}
92 if(search(c,hc)&&search(d,hd)) {ans++;}
93 }
94 }
95 out(ans/4);putchar(‘\n‘);
96 }
97 }
原文地址：https://www.cnblogs.com/codeoosacm/p/9822855.html