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.
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: "); 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 ",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 ",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(' '); 96 } 97 }