POJ 2002 Squares [hash]

时间：2014-12-12 18:37:45 阅读：147 评论：0 收藏：0 [点我收藏+]

标签：des style blog http io ar color os sp

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 16631		Accepted: 6328

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

Sample Output

1
6
1

Source

Rocky Mountain 2004

题解：枚举两个点，算出另外两点坐标，看是否在给的点集里。

具体实现用hash，hash碰撞了就放在链表中，然后在链表里查找~（天猫所说的 pascal拉链哈希，，，

就写了个哈希函数然后对函数值指针拉链出来哈希）

）

  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<map>
  6 #define ll long long
  7 #define N 1005
  8 #define mod 2007
  9 
 10 using namespace std;
 11 
 12 int n;
 13 int x[2*N];
 14 int y[2*N];
 15 //map<pair<int,int>,int>mp;
 16 int ans;
 17 int head[2*N];
 18 int next[2*N];
 19 int m;
 20 
 21 void insert(int i)
 22 {
 23     int key=(x[i]*x[i]+y[i]*y[i])%mod;
 24     next[m]=head[key];
 25     x[m]=x[i];
 26     y[m]=y[i];
 27     head[key]=m++;
 28 }
 29 
 30 void ini()
 31 {
 32     ans=0;
 33     m=N;
 34     memset(head,-1,sizeof(head));
 35     //mp.clear();
 36     int i;
 37     for(i=1;i<=n;i++){
 38         scanf("%d%d",&x[i],&y[i]);
 39         insert(i);
 40         //mp[ make_pair(x[i],y[i]) ]=i;
 41     }
 42 }
 43 
 44 int find(int xx,int yy)
 45 {
 46     int key=(xx*xx+yy*yy)%mod;
 47     int i;
 48     for(i=head[key];i!=-1;i=next[i]){
 49         if(x[i]==xx && y[i]==yy){
 50             return 1;
 51         }
 52     }
 53     return 0;
 54 }
 55 
 56 int ok(int i,int j)
 57 {
 58     int mx,my;
 59     int x3,x4,y3,y4;
 60     int he,cha;
 61     mx=x[i]+x[j];
 62     my=y[i]+y[j];
 63     he=mx+my;cha=my-mx;
 64     if(he%2!=0 || cha%2!=0) return 0;
 65     he/=2;cha/=2;
 66     x3=he-y[i];
 67    // x3=mx+my-y[i];
 68     y3=cha+x[i];
 69  //   y3=my-(mx-x[i]);
 70     if(find(x3,y3)==0) return 0;
 71     x4=y[i]-cha;
 72    // x4=mx-(my-y[i]);
 73    // y4=my+(mx-x[i]);
 74     y4=he-x[i];
 75     if(find(x4,y4)==0) return 0;
 76 
 77     return 1;
 78 }
 79 
 80 void solve()
 81 {
 82     int i,j;
 83     for(i=1;i<n;i++){
 84         for(j=i+1;j<=n;j++){
 85             if(ok(i,j)==1){
 86                 ans++;
 87             }
 88         }
 89     }
 90 }
 91 
 92 void out()
 93 {
 94     ans/=2;
 95     printf("%d\n",ans);
 96 }
 97 
 98 int main()
 99 {
100     //freopen("data.in","r",stdin);
101    // scanf("%d",&T);
102     //while(T--){
103     while(scanf("%d",&n)!=EOF){
104         if(n==0) break;
105         ini();
106         solve();
107         out();
108     }
109     return 0;
110 }

POJ 2002 Squares [hash]

标签：des style blog http io ar color os sp

原文地址：http://www.cnblogs.com/njczy2010/p/4160159.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行