hdoj--5563--Clarke and five-pointed star（简单几何）

Clarke and five-pointed star

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 601 Accepted Submission(s): 322

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.
)

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

Yes
No

Hint

[align=left]Source[/align]
BestCoder Round #62 (div.2)

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五角星是一个对称性极高的图形，每一个点连得线都与其他点的连线一毛一样，有木有很神奇（并没有），这道题只需要算出每一个点的所有连线，然后判断就好，水题一枚

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct node
{
double x,y;
}p[10010];
double dis1[10],dis2[10],dis3[10],dis4[10],dis5[10];
double d(node s1,node s2)
{
return sqrt((s1.x-s2.x)*(s1.x-s2.x)+(s1.y-s2.y)*(s1.y-s2.y));
}
int cmp(double a,double b)
{
if(a>b)
return 1;
return 0;
}
int er(double a,double b)
{
if(fabs(a-b)<1e-3)
return 1;
return 0;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int i,j;
for(i=0;i<5;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
for(j=0;j<5;j++)
dis1[j]=d(p[0],p[j]);
for(j=0;j<5;j++)
dis2[j]=d(p[1],p[j]);
for(j=0;j<5;j++)
dis3[j]=d(p[2],p[j]);
for(j=0;j<5;j++)
dis4[j]=d(p[3],p[j]);
for(j=0;j<5;j++)
dis5[j]=d(p[4],p[j]);
sort(dis1,dis1+5,cmp);
sort(dis2,dis2+5,cmp);
sort(dis3,dis3+5,cmp);
sort(dis4,dis4+5,cmp);
sort(dis5,dis5+5,cmp);
for(i=0;i<5;i++)
{
if(er(dis1[i],dis2[i])&&
er(dis2[i],dis3[i])&&
er(dis3[i],dis4[i])&&
er(dis4[i],dis5[i])&&
er(dis5[i],dis1[i]));
else break;
}
if(i==5)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}