HDOJ 5563 Clarke and five-pointed star (判断五个点组成的是否为正五角星)

Clarke and five-pointed star

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 420    Accepted Submission(s): 227

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 than10−4.
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

23.0000000 0.00000000.9270509 2.85316950.9270509 -2.8531695-2.4270509 1.7633557-2.4270509 -1.76335573.0000000 1.00000000.9270509 2.85316950.9270509 -2.8531695-2.4270509 1.7633557-2.4270509 -1.7633557

 

Sample Output

YesNo
Hint
题意：给你5个点，求这5个点能不能连成一个正五角星思路：见每个点都连起来会发现，会组成一个五边形，而正五角星组成的是正五边形，所一说此题就转化为判断是否能组成正五边形直接枚举距离，正多边形个点之间的连线中边长一定是最短的那条边，所以我们只需要查看最短边数量是否等于5就可以判断是不是正五边形，也就能判断能不能组成正五角星。需要注意的是精度问题。可能这道题数据有点弱，窝水了过去，正式的应该是先求凸包，然后在判断，但是可能因为精度问题水了过去。。类似的题有：点击打开链接ac代码： 
<pre name="code" class="cpp">#include<stdio.h>#include<string.h>#include<math.h>#include<stack>#include<iostream>#include<algorithm>#define fab(a) (a)>0?(a):(-a)#define LL long long#define MAXN 1000010#define mem(x) memset(x,0,sizeof(x))#define INF 0xfffffff using namespace std;struct s{double x,y;}a[10];int v[10][10];double dis[1000];double fun(s aa,s bb){return sqrt((aa.x-bb.x)*(aa.x-bb.x)+(aa.y-bb.y)*(aa.y-bb.y));}int main(){int t,i,j;scanf("%d",&t);while(t--){for(i=0;i<5;i++)scanf("%lf%lf",&a[i].x,&a[i].y);mem(v);int k=0;double mi=1.0*INF;for(i=0;i<5;i++){for(j=i+1;j<5;j++){if(v[i][j]||v[j][i])continue;dis[k]=fun(a[i],a[j]);mi=min(mi,dis[k]);v[i][j]=1; k++;}}int cnt=0;for(i=0;i<k;i++){if(dis[i]-mi<=1e-4)//精度问题cnt++;}if(cnt==5)printf("Yes\n");elseprintf("No\n");}return 0;}