HDU 5563 Clarke and five-pointed star(判断正五边形)——BestCoder Round #62(div.1 div.2)
Clarke and five-pointed star
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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.
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.
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 . )
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 NoHint
Source
BestCoder Round #62 (div.2)
/************************************************************************/
附上该题对应的中文题
Clarke and five-pointed star
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
问题描述
克拉克是一名人格分裂患者。某一天克拉克分裂为一个几何学习者，在研究多边形。 在研究某一个多边形的时候，克拉克发现他多次遇到判断5个点是否能组成一个五角星的问题，在这里，这5个点分别代表五角星的五个顶点（顶角上的点）。于是他跑来想你求助，让你写出一个程序快速判定。即对于给出的5个点，判断这5个点是否能组成一个五角星。
输入描述
第一行一个整数T(1≤T≤10)，表示数据的组数。 每组数据有5行，每行有两个实数xi,yi(−109≤xi,yi≤109)，表示第i个点的坐标。
输出描述
如果两个量相差小于10−4，则认为这两个量相等。 对于每组数据，如果这5个点能组成一个五角星，则输出Yes，否则输出No。（如果5个点相同，那么也能组成一个五角星。）
输入样例
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
输出样例
Yes No
Hint
样例1如图
/****************************************************/
出题人的解题思路：
Clarke and five-pointed star
容易看出只需要判断这5个点是否在一个正五边形上。
因此我们枚举排列，然后依次判断即可。
判定方法是，五条相邻边相等，五条对角线相等。
当然题目给的精度问题，窝只能说，如果泥做法不复杂，精度足够好的话，是可以过的。毕竟题目说的小于10−4是指理论上的，所以理论上适用所有的数之间的比较。所以有人问我开方前和开方后，我只能说，哪个精度高用哪个....
当然你也可以先求出凸包然后再判相邻距离......
这题明面上是让你判断5个顶点能否构成五角星，但是考虑到五角星的顶点与其对应的正五边形的顶点一一对应，故此题就转化成了判断5个给定的顶点能否构成正五边形
然而并不会证明正五边形，所以把所有点两两之间的边都求出来，排序后，判等，又因为担心这样还不够，还多求了一下角度等于108°
另外，题目的精度要注意， 10−4 ，如果不控制精度，求边开方四舍五入后会造成误差
10−410−4
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-4
using namespace std;
const int N = 5;
const int M = 100005;
const int inf = 100000000;
const int mod = 2009;
const double PI = acos(-1);
struct node
{
double x,y;
}s[N];
double Abs(double x)
{
if(x>0)
return x;
return -x;
}
double ans[2*N],z;
int main()
{
int t,i,j,k;
scanf("%d",&t);
while(t--)
{
for(i=0;i<5;i++)
scanf("%lf%lf",&s[i].x,&s[i].y);
for(k=i=0;i<5;i++)
for(j=0;j<i;j++)
ans[k++]=sqrt((s[i].x-s[j].x)*(s[i].x-s[j].x)+(s[i].y-s[j].y)*(s[i].y-s[j].y));
sort(ans,ans+10);//printf("%lf\n",PI);
z=180*acos((ans[0]*ans[0]+ans[1]*ans[1]-ans[9]*ans[9])/ans[0]/ans[1]/2)/PI;//printf("%lf %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf\n",ans[0],ans[1],ans[2],ans[3],ans[4],ans[5],ans[6],ans[7],ans[8],ans[9],z);
if(Abs(ans[0]-ans[1])<exp&&Abs(ans[1]-ans[2])<exp&&Abs(ans[2]-ans[3])<exp&&Abs(ans[3]-ans[4])<exp&&Abs(ans[4]-ans[5])>exp&&Abs(ans[5]-ans[6])<exp&&Abs(ans[6]-ans[7])<exp&&Abs(ans[7]-ans[8])<exp&&Abs(ans[8]-ans[9])<exp&&Abs(z-108)<exp)
puts("Yes");
else
puts("No");
}
return 0;
}
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