Rectangle and Square

最新推荐文章于 2019-10-14 22:20:03 发布

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最新推荐文章于 2019-10-14 22:20:03 发布

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原文链接：http://www.cnblogs.com/arno-my-boke/p/4738472.html

本文介绍了一种利用C++ STL库中的全排列函数解决特定几何问题的方法，具体涉及将一组点分为两组，形成一个正方形和一个矩形。通过代码示例详细解释了解题过程。

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Description

Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.

Input

You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 10⁴. It is guaranteed that no tw

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
using namespace std;
int p[]={0,1,2,3,4,5,6,7,8};
int x[9],y[9];
int d(int a,int b)
{
    return (x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b]);
}
int main()
{
    int i,j;
    for(i=1;i<=8;i++)scanf("%d%d",&x[i],&y[i]);

    do
    {
        if(d(p[1],p[2])==d(p[2],p[3]) && d(p[2],p[3])==d(p[3],p[4]) && d(p[3],p[4])==d(p[4],p[1]) && d(p[1],p[3])==d(p[4],p[2])
           && d(p[5],p[6])==d(p[7],p[8]) && d(p[5],p[8])==d(p[6],p[7]) && d(p[5],p[7])==d(p[6],p[8]))
           {
               return printf("YES\n%d %d %d %d\n%d %d %d %d\n",p[1],p[2],p[3],p[4],p[5],p[6],p[7],p[8]),0;
           }
    }while(next_permutation(p+1,p+9));
   printf("NO\n"); 
    return 0;




}

o points coincide.

Output

Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.

If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.

Sample Input

Input

Output

YES
5 6 7 8
1 2 3 4

Input

Output

NO

Input

Output

YES
1 2 3 4
5 6 7 8
一开始不想敲，认为就是个全排列，没什么，网上搜了一下解题报告，确实是用全排列把所有的情况找出来，但是别人用了一个我认为非常吊的方法，调用stl里的全排列函数，屌爆了。
膜拜一下，STL里还有什么宝贝赶紧去找找。

转载于:https://www.cnblogs.com/arno-my-boke/p/4738472.html