hdu 5563 Clarke and five-pointed star (水题)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 351    Accepted Submission(s): 197

 

 

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

 

 

 

 

 

 

遍历五个点，找到与当前点最远的点，求出距离。可以求得五个距离，如果这五个距离相等，则可以组成一个正五角星。

 

 

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define PI acos(-1)

int cmp(const void *a, const void *b){
　　return *(double *)a > *(double *)b ? 1 : -1;
}
int main(){
　　int T;
　　scanf("%d", &T);
　　while(T--){
　　　　double x[5], y[5];
　　　　int i, j;
　　　　int count = 0;
　　　　double hehe, hoho;
　　　　for(i = 0; i < 5; i++) scanf("%lf%lf", &x[i], &y[i]); //保存下五个点的数据
　　　　for(i = 0; i < 5; i++){
　　　　　　double dis[5]; //操作当前点
　　　　　　for(j = 0; j < 5; j++){
　　　　　　　　dis[j] = sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));
　　　　　　}
　　　　qsort(dis, 5, sizeof(double), cmp); //排序找到距离当前点最远的点
　　　　double dismax;
　　　　dismax = dis[4];
　　　　if(count == 0) hehe = dismax;
　　　　if(-0.0001 < hehe - dismax && hehe - dismax < 0.0001) count++;
　　　　hehe = dismax;
　　　　}

　　　　if(count == 5) printf("Yes\n"); //五条边相等
　　　　else printf("No\n");
　　}
　　return 0;
}