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Problem DescriptionFat brother and Maze are playing a kind of special (hentai) game in the clearly blue sky which we can just consider as a kind of two-dimensional plane. Then Fat brother starts to draw N starts in the sky which we can just consider each as a point. After he draws these stars, he starts to sing the famous song “The Moon Represents My Heart” to Maze.
You ask me how deeply I love you,
How much I love you?
My heart is true,
My love is true,
The moon represents my heart.
…
But as Fat brother is a little bit stay-adorable(呆萌), he just consider that the moon is a special kind of convex quadrilateral and starts to count the number of different convex quadrilateral in the sky. As this number is quiet large, he asks for your help.
InputThe first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each case contains an integer N describe the number of the points.
Then N lines follow, Each line contains two integers describe the coordinate of the point, you can assume that no two points lie in a same coordinate and no three points lie in a same line. The coordinate of the point is in the range[-10086,10086].
1 <= T <=100, 1 <= N <= 30
OutputFor each case, output the case number first, and then output the number of different convex quadrilateral in the sky. Two convex quadrilaterals are considered different if they lie in the different position in the sky.
Sample Input Sample Output//140 ms 216KB
#include<stdio.h>
#include<math.h>
#define eps 1e-9
struct P
{
int x,y;
}p[1007];
double area(P a, P b, P c)
{
return a.x * b.y + c.x * a.y + b.x * c.y - c.x * b.y - a.x * c.y - b.x * a.y;
}
bool ok(P a,P b,P c,P d)
{
double sum1 = fabs(area(a, b, d)) + fabs(area(a, c, d)) + fabs(area(b, c, d));
double sum2 = fabs(area(a, b, c));
if(sum1==sum2)return true;
return false;
}
bool judge(P a,P b,P c,P d)
{
if(ok(a,b,c,d))return false;
if(ok(a,b,d,c))return false;
if(ok(a,c,d,b))return false;
if(ok(b,c,d,a))return false;
return true;
}
int main()
{
int t,cas=1;
scanf("%d",&t);
while(t--)
{
int n,ans=0;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d%d",&p[i].x,&p[i].y);
for(int i=0;i<n;i++)
for(int j=i+1;j<n;j++)
for(int x=j+1;x<n;x++)
for(int y=x+1;y<n;y++)
if(judge(p[i],p[j],p[x],p[y]))
ans++;
printf("Case %d: %d\n",cas++,ans);
}
return 0;
}
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原文地址：http://blog.csdn.net/crescent__moon/article/details/45103221
