HDU 5562 Clarke and five-pointed star (BC)

本文介绍了一种通过坐标判断五个点是否能构成正五角星的算法。提供了两种方法：一是将点按逆时针排序并检查边长；二是计算所有点间距离，对比边长与对角线长度。

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Clarke and five-pointed star

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/65536K (Java/Other)

Total Submission(s) : 8 Accepted Submission(s) : 2

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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)

分析：给出5个点的坐标，判断是否是一个正五角星

方法一：判断是正五角星，只需要判断是正五边形，因为输入的点是随机输入的，所以先把点按逆时针排一下序，然后再计算五边形的边是否相等即可。

方法二：把每个点到其他各点的距离求出来，排序，小的五条边是五边形的边，大的五条边是五边形的对角线。判断小的边相等，大的边相等即可。

代码如下（排序）：

#include <stdio.h>
#include <math.h>
typedef struct node{
	double x,y;
	double k;
}Node;
Node ans[10],left[10],right[10];
double sumx,sumy;
void sort()
{//点的排序 
	int i,j;
	sumx=sumx/5.0;
	sumy=sumy/5.0;
	int le,ri;
	le=ri=0;
	for(i=0;i<5;i++)
	{
		ans[i].k=(ans[i].y-sumy)/(ans[i].x-sumx);
		if(ans[i].x<=sumx)
			left[le++]=ans[i];
		else
			right[ri++]=ans[i];
	}
	for(i=0;i<le-1;i++)
	{
		for(j=i+1;j<le;j++)
		{
			if(left[i].k > left[j].k)
			{
				Node temp=left[i];
				left[i]=left[j];
				left[j]=temp;
			}
		}
	}
	for(i=0;i<ri-1;i++)
	{
		for(j=i+1;j<ri;j++)
		{
			if(right[i].k>right[j].k)
			{
				Node temp=right[i];
				right[i]=right[j];
				right[j]=temp;
			}
		}
	}
	int num=0;
	for(i=0;i<le;i++)
		ans[num++]=left[i];
	for(i=0;i<ri;i++)
		ans[num++]=right[i];
}

int main()
{
	int T;
	int i; 
	scanf("%d",&T);
	while(T--)
	{
		sumx=sumy=.0;
		for(i=0;i<5;i++)
		{
			scanf("%lf %lf",&ans[i].x,&ans[i].y);
			sumx+=ans[i].x;
			sumy+=ans[i].y;
		}
		sort();
		double avg = sqrt((ans[0].x-ans[1].x)*(ans[0].x-ans[1].x)+(ans[0].y-ans[1].y)*(ans[0].y-ans[1].y));
		int peace=0;
		for(i=1;i<4;i++)
		{
			double temp=sqrt((ans[i].x-ans[i+1].x)*(ans[i].x-ans[i+1].x)+(ans[i].y-ans[i+1].y)*(ans[i].y-ans[i+1].y));
			if(fabs(avg - temp)<=0.0001)
				continue;
			else{
				peace=1;
				break;
			}
		}
		if(!peace)
			printf("Yes\n");
		else
			printf("No\n");
	}
	
	return 0;
}