FZU Moon Game(几何)

Accept: 710    Submit: 2038

Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

Fat 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.

 Input

The 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

 Output

For 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

240 0100 00 100100 10040 0100 00 10010 10

 Sample Output

Case 1: 1Case 2: 0

我直接用了凸包算法，大材小用了

#include <iostream>
#include <string.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>

using namespace std;
struct Node
{
    int x;
    int y;
}a[5],b[35];
int s[10];
int top;
int cross(Node a,Node b,Node c)
{
    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}
int dis(Node a,Node b)
{
    return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int cmp(Node p1,Node p2)
{
    int temp=cross(a[0],p1,p2);
    if(temp>0) return true;
    else if(temp==0&&dis(a[0],p1)<dis(a[0],p2)) return true;
    else return false;
}
int cmp2(Node a,Node b)
{
    if(a.y==b.y)
        return a.x<b.x;
    return a.y<b.y;
}
int graham(int n)
{
    if(n==1){return 0;}
    if(n==2){return 1;}
    if(n>2)
    {
        top=1;s[0]=0;s[1]=1;
        for(int i=2;i<n;i++)
        {
            while(top>0&&cross(a[s[top-1]],a[s[top]],a[i])<=0)
                top--;
            s[++top]=i;
        }

        return top;

    }

}
int n;
int main()
{
    int t;
    scanf("%d",&t);
    int cas=0;
    int ans=0;
    while(t--)
    {
       scanf("%d",&n);
       ans=0;
       for(int i=1;i<=n;i++)
            scanf("%d%d",&b[i].x,&b[i].y);
       for(int i=1;i<=n;i++)
       {

        for(int j=i+1;j<=n;j++)
        {

          for(int k=j+1;k<=n;k++)
          {

            for(int p=k+1;p<=n;p++)
            {
                a[0]=b[i];a[1]=b[j];a[2]=b[k];a[3]=b[p];
                sort(a,a+4,cmp2);

                sort(a+1,a+4,cmp);

                //cout<<i<<" "<<j<<" "<<k<<" "<<p<<endl;
                if(graham(4)==3)
                {
                   // cout<<i<<" "<<j<<" "<<k<<" "<<p<<endl;
                    ans++;
                }
            }
          }
        }
       }
       printf("Case %d: %d\n",++cas,ans);
    }
    return 0;
}