• poj 2002 -- Squares


    Squares
    Time Limit: 3500MS   Memory Limit: 65536K
    Total Submissions: 15886   Accepted: 6013

    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
    
    思路:
        正方形已知两点(x1,y1),(x2,y2)求另两点坐标:
          x3=x1+y1-y2;
          y3=y1-x1+x2;
          x4=x2+y1-y2;
          y4=y2-x1+x2;

          x3=x1-y1+y2;
          y3=y1+x1-x2;
          x4=x2-y1+y2;
          y4=y2+x1-x2;
    计算出坐标后可以用hash查找。用静态邻接表模拟拉链。。

     1 /*======================================================================
     2  *           Author :   kevin
     3  *         Filename :   Squares.cpp
     4  *       Creat time :   2014-07-25 09:41
     5  *      Description :
     6 ========================================================================*/
     7 #include <iostream>
     8 #include <algorithm>
     9 #include <cstdio>
    10 #include <cstring>
    11 #include <queue>
    12 #include <cmath>
    13 #define clr(a,b) memset(a,b,sizeof(a))
    14 #define M 20000
    15 using namespace std;
    16 struct Node
    17 {
    18     int x,y;
    19 }node[1005];
    20 int head[2*M+5];
    21 struct EdgeNode{
    22     int to,next;
    23 };
    24 EdgeNode Edges[2*M+5];
    25 
    26 void AddEdges(int i,int j,int cnt)
    27 {
    28     Edges[cnt].to = j;
    29     Edges[cnt].next = head[i];
    30     head[i] = cnt;
    31 }
    32 bool judge(int x,int y)
    33 {
    34     for(int k = head[x]; k != -1; k = Edges[k].next){
    35         if(Edges[k].to == y)
    36             return true;
    37     }
    38     return false;
    39 }
    40 int main(int argc,char *argv[])
    41 {
    42     int n;
    43     while(scanf("%d",&n)!=EOF && n){
    44         clr(node,0);
    45         clr(head,-1);
    46         clr(Edges,0);
    47         int k = 0;
    48         for(int i = 0; i < n; i++){
    49             scanf("%d%d",&node[i].x,&node[i].y);
    50             AddEdges(M+node[i].x,M+node[i].y,k++);
    51         }
    52         int x1,x2,x3,x4,y1,y2,y3,y4;
    53         int cnt = 0;
    54         for(int i = 0; i < n; i++){
    55             for(int j = i+1; j < n; j++){
    56                 x1 = node[i].x + node[i].y - node[j].y;
    57                 y1 = node[i].y - node[i].x + node[j].x;
    58                 x2 = node[j].x + node[i].y - node[j].y;
    59                 y2 = node[j].y - node[i].x + node[j].x;
    60                 if(judge(x1+M,y1+M)){
    61                     if(judge(x2+M,y2+M)){
    62                         cnt++;
    63                     }
    64                 }
    65                 x3 = node[i].x - node[i].y + node[j].y;
    66                 y3 = node[i].y + node[i].x - node[j].x;
    67                 x4 = node[j].x - node[i].y + node[j].y;
    68                 y4 = node[j].y + node[i].x - node[j].x;
    69                 if(judge(x3+M,y3+M)){
    70                     if(judge(x4+M,y4+M)){
    71                         cnt++;
    72                     }
    73                 }
    74             }
    75         }
    76         printf("%d
    ",cnt/4);
    77     }
    78     return 0;
    79 }
    View Code
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  • 原文地址:https://www.cnblogs.com/ubuntu-kevin/p/3867398.html
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