Friday, December 2, 2016

HDU 5365 Run - BestCoder Round #50(div.1 div.2)


http://acm.hdu.edu.cn/showproblem.php?pid=5365
AFA is a girl who like runing.Today,she downloaded an app about running .The app can record the trace of her runing.AFA will start runing in the park.There are many chairs in the park,and AFA will start her running in a chair and end in this chair.Between two chairs,she running in a line.she want the the trace can be a regular triangle or a square or a regular pentagon or a regular hexagon.
Please tell her how many ways she can find.
Two ways are same if the set of chair that they contains are same.
Input
There are multiply case.
In each case,there is a integer n(1 < = n < = 20)in a line.
In next n lines,there are two integers xi,yi(0 < = xi,yi < 9) in each line.
Output
Output the number of ways.
Sample Input
4 0 0 0 1 1 0 1 1
Sample Output
1

至于判断正方形,只要判断四边相等和一条对角线是边长的根号2倍就行了。
http://blog.csdn.net/queuelovestack/article/details/47364285
地球人都知道整点是不能构成正五边形和正三边形和正六边形的,所以只需暴力枚举四个点判断是否是正四边形即可。假如你不是地球人,那么即使暴力枚举正三边形和稍微不那么暴力地找正五边形和正六边形也是可以通过的(反正找不到)。
出题人简单粗暴地直接说明了整点是不能构成正三角形、正五边形和正六边形的,相信比较多的人对此都有疑问,为什么不能构成呢,下面我给出不能构成正三角形的证明
反证法:假设存在坐标平面上的三个整点,组成了正三角形
则将其平移,将其一个顶点与坐标原点重合,则另两个顶点仍然是整点
设一个非原点的顶点M坐标为(a,b) ,另一个顶点为N(c,d), a、b、c、d都是整数
则MO的斜率=,倾斜角=arctan()
则NO的倾斜角=arctan()+
NO的斜率=
因为a、b不全等于0,所以c和d至少有一个是无理数
这和c、d都是整数矛盾
所以坐标平面上的三个整点,一定不能组成正三角形
至于正五边形和正六边形,就交给你们自己去证明了。所以此题只要暴力枚举每四个点,判断能否形成正方形即可,而正方形的判断依据是四条边均相等,对角线相等,且其平方是边平方的两倍。
  1. const int N = 25;  
  2. const int inf = 1000000000;  
  3. const int mod = 1000000007;  
  4. int x[N],y[N],s[6];  
  5. int sqare(int x1,int y1,int x2,int y2)  
  6. {  
  7.     return (x2-x1)*(x2-x1)+(y2-y1)*(y2-y1);  
  8. }  
  9. bool judge(int i,int j,int k,int l)  
  10. {  
  11.     int t=0;  
  12.     s[t++]=sqare(x[i],y[i],x[j],y[j]);  
  13.     s[t++]=sqare(x[j],y[j],x[k],y[k]);  
  14.     s[t++]=sqare(x[k],y[k],x[l],y[l]);  
  15.     s[t++]=sqare(x[l],y[l],x[i],y[i]);  
  16.     s[t++]=sqare(x[i],y[i],x[k],y[k]);  
  17.     s[t++]=sqare(x[l],y[l],x[j],y[j]);  
  18.     sort(s,s+6);  
  19.     if(s[0]==s[1]&&s[2]==s[3]&&s[1]==s[2]&&s[4]==2*s[0]&&s[5]==s[4])  
  20.         return true;  
  21.     return false;  
  22. }  
  23. int main()  
  24. {  
  25.     int n,i,j,k,l,t;  
  26.     while(~scanf("%d",&n))  
  27.     {  
  28.         t=0;  
  29.         for(i=1;i<=n;i++)  
  30.             scanf("%d%d",&x[i],&y[i]);  
  31.         for(i=1;i<=n;i++)  
  32.             for(j=i+1;j<=n;j++)  
  33.                 for(k=j+1;k<=n;k++)  
  34.                     for(l=k+1;l<=n;l++)  
  35.                         if(judge(i,j,k,l))  
  36.                             t++;  
  37.         printf("%d\n",t);  
  38.     }  
  39.     return 0;  
http://www.cnblogs.com/Emerald/p/4720631.html


http://www.voidcn.com/blog/aaaaacmer/article/p-4310045.html
题目大意:给定一些整数点,问这些点能够组成多少个正三角形或者是正方形、正五边形、正六边形。
思路:如果仔细想一想的话,这道题目是根本不存在正三角形、正五边形、正六边形的请款的。(在纸上画图看一看,可以发现确实不行。题目输入是整数点)
所以题目变成:这些点能够组成多少个正方形。看数据范围,n<=20,坐标范围<=8。
题目瞬间简单了,暴力!暴力枚举4个点,然后判断这4个点是否能组成一个正方形。判断是否为正方形的方法很多,向量或者是距离。
我这里用的是算出4个点的距离,还有2个对角线的距离。如果能组成正方形,条件是:4条边长度相同,2条对角线长度相同。


https://my.oschina.net/sunshinedabby/blog/57529
    在做ACM题目的时候,遇到了要求输入四个点的坐标,然后判断这四个点能否构成一个正方形.这个看似很简单,但细节方面还是有许多问题的.我想的方法很容易理解,而且一次就AC了.
    首先,我先选定三个点,判断这三个点是否可以构成一个等腰直角三角形(如果不是的话,肯定无法构成正方形的),在判断的同时,还要确定一下三个点(主要是直角的那个点);这样确定后,再利用第四个点验证一下就容易了.这题要求的是穷举法,我想在万不得已的情况下才会考虑这样的方法的.不过,作为初学者练习穷举法的使用技巧还是很推荐使用.
bool Judge(Coordinate const x,Coordinate const y,Coordinate const z)
{
    Coordinate *mVector = new Coordinate(x.xCoordinate - y.xCoordinate,
x.yCoordinate - y.yCoordinate);
    Coordinate *nVector = new Coordinate(z.xCoordinate - 
(x.xCoordinate + y.xCoordinate)/2,
                                         z.yCoordinate - 
                                         (x.yCoordinate + y.yCoordinate)/2);
    //判断是否为等腰三角形
    bool result = ((mVector->xCoordinate * nVector->xCoordinate +
                    mVector->yCoordinate * nVector->yCoordinate) == 0);

    //判断是否是直角三角形
    if(result)
        result = (mVector->xCoordinate * mVector->xCoordinate + 
mVector->yCoordinate * mVector->yCoordinate)
                 == ((nVector->xCoordinate * nVector->xCoordinate + 
nVector->yCoordinate * nVector->yCoordinate) * 4);
                 
    delete mVector;
    delete nVector;
    
    return result;
}

/**
 * @explanation 给定四个坐标点,判断这四个坐标点是否为一个正方形上的四点.
 * @algorithm  任意选定三点,判断是否为等腰直角三角形(确定直角的位置);
 * 然后再利用另外一个点来验证.
 */
bool IsSquare(Coordinate *array,int length)
{
    if(length != 4)
        return false;
    int a,b,c;

    if(Judge(array[0],array[1],array[2]))
    {
        a = 0;
        b = 1;
        c = 2;
    }
    else if(Judge(array[0],array[2],array[1]))
    {
        a = 0;
        b = 2;
        c = 1;
    }
    else if(Judge(array[2],array[1],array[0]))
    {
        a = 1;
        b = 2;
        c = 0;
    }
    else
        return false;

    return (array[3] != array[c] && Judge(array[a],array[b],array[3]));
}


http://blog.csdn.net/u011394362/article/details/40141929

如何证明平面内任意六个整点都不能组成正六边形?
https://www.zhihu.com/question/25304120?rf=34324133

http://www.1point3acres.com/bbs/thread-210789-1-1.html
又出了第二道。给定四个二维坐标,判断这四个点是否能组成一个正方形。这题在跑test case的时候意识到出现点小问题,没时间改了,就跟面试官说了思路。

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