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  • hdoj--5333--Dancing Stars on Me(水题)

    Dancing Stars on Me

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
    Total Submission(s): 582    Accepted Submission(s): 308


    Problem Description
    The sky was brushed clean by the wind and the stars were cold in a black sky. What a wonderful night. You observed that, sometimes the stars can form a regular polygon in the sky if we connect them properly. You want to record these moments by your smart camera. Of course, you cannot stay awake all night for capturing. So you decide to write a program running on the smart camera to check whether the stars can form a regular polygon and capture these moments automatically.

    Formally, a regular polygon is a convex polygon whose angles are all equal and all its sides have the same length. The area of a regular polygon must be nonzero. We say the stars can form a regular polygon if they are exactly the vertices of some regular polygon. To simplify the problem, we project the sky to a two-dimensional plane here, and you just need to check whether the stars can form a regular polygon in this plane.
     

    Input
    The first line contains a integer T indicating the total number of test cases. Each test case begins with an integer n, denoting the number of stars in the sky. Following n lines, each contains 2 integers xi,yi, describe the coordinates of n stars.

    1T300
    3n100
    10000xi,yi10000
    All coordinates are distinct.
     

    Output
    For each test case, please output "`YES`" if the stars can form a regular polygon. Otherwise, output "`NO`" (both without quotes).
     

    Sample Input
    3 3 0 0 1 1 1 0 4 0 0 0 1 1 0 1 1 5 0 0 0 1 0 2 2 2 2 0
     

    Sample Output
    NO YES NO



    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #include<math.h>
    using namespace std;
    #define MAX 100010
    struct node
    {
    	int x,y;
    }point[MAX];
    int map[1010][1010];
    double dis[MAX];
    double dist(node s1,node s2)
    {
    	double s=sqrt((1.0*s1.x-s2.x)*(s1.x-s2.x)+(s1.y-s2.y)*(s1.y-s2.y));
    	return s;
    }
    int main()
    {
    	int t;
    	scanf("%d",&t);
    	while(t--)
    	{
    		int n;
    		double minn=10000000;
    		memset(dis,0,sizeof(dis));
    		memset(map,0,sizeof(map));
    		scanf("%d",&n);
    		for(int i=0;i<n;i++)
    		scanf("%d%d",&point[i].x,&point[i].y);
    		int cnt=0;
    		for(int i=0;i<n;i++)
    		for(int j=0;j<n;j++)
    		{
    			if(i==j) continue;
    			if(!map[i][j])
    			{
    				dis[cnt]=dist(point[i],point[j]);
    				if(minn>dis[cnt]&&dis[cnt]!=0) 
    				minn=dis[cnt];
    				cnt++;
    				map[i][j]=1;
    			}
    		}
    		int ans=0;
    		for(int i=0;i<cnt;i++)
    		if(minn==dis[i]) ans++;
    		//printf("%d %d %d",ans,n,minn);
    		if(ans==2*n)
    		printf("YES
    ");
    		else
    		printf("NO
    ");
    	}
    	return 0;
    }

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  • 原文地址:https://www.cnblogs.com/playboy307/p/5273638.html
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