HDOJ 5563 Clarke and five-pointed star

Clarke and five-pointed star

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 865    Accepted Submission(s): 454

Problem Description

Clarke is a patient with multiple personality disorder. One day, Clarke turned into a learner of geometric. 
When he did a research with polygons, he found he has to judge if the polygon is a five-pointed star at many times. There are 5 points on a plane, he wants to know if a five-pointed star existed with 5 points given.

 

Input

The first line contains an integer T(1≤T≤10), the number of the test cases. 
For each test case, 5 lines follow. Each line contains 2 real numbers xi,yi(−109≤xi,yi≤109), denoting the coordinate of this point.

 

Output

Two numbers are equal if and only if the difference between them is less than 10−4. 
For each test case, print Yes if they can compose a five-pointed star. Otherwise, print No. (If 5 points are the same, print Yes. )

 

Sample Input

2 3.0000000 0.0000000 0.9270509 2.8531695 0.9270509 -2.8531695 -2.4270509 1.7633557 -2.4270509 -1.7633557 3.0000000 1.0000000 0.9270509 2.8531695 0.9270509 -2.8531695 -2.4270509 1.7633557 -2.4270509 -1.7633557

 

Sample Output

Yes No

Hint

 

Source

BestCoder Round #62 (div.2)

 

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     題意就是你5個頂點，判斷這五個頂點是否能組成五角星；

    一個判斷就是這五個點與其他四個點的距離是相等的，依次計算然後判斷就行了，這裏要注意的就是精度問題，如果兩個量相差小於10的-4次方，則認爲這兩個量相等。

   所以判斷的時候要注意精度。

#include<stdio.h>
#include<algorithm>
#include<string>
#include<string.h>
#include<queue>
#include<math.h>
using namespace std;
struct p
{
    double x;
    double y;
    double len[4];
}a[5];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=0;i<5;i++)
            scanf("%lf%lf",&a[i].x,&a[i].y);
        for(int i=0;i<5;i++)
        {
            int l=0;
           for(int j=0;j<5;j++)
           {
               if(i==j)
                continue;
               a[i].len[l++]=sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));
           }
        }
        for(int i=0;i<5;i++)
            sort(a[i].len,a[i].len+4);
        int f=0;
        for(int i=0;i<5;i++)
        {
            for(int j=0;j<5;j++)
            {
                for(int k=0;k<4;k++)
                {
                    if((a[i].len[k]-a[j].len[k])>0.00001)
                    {
                        printf("No\n");
                        f=1;
                        break;
                    }
                }
                if(f==1)
                    break;
            }
            if(f==1)
                break;
        }
        if(f==1)
            continue;
        printf("Yes\n");
    }
return 0;
}