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
2 4 0 0 100 0 0 100 100 100 4 0 0 100 0 0 100 10 10
Sample Output
Case 1: 1 Case 2: 0
/*/
这个题目也是比赛的题目，虽然没这么难，但是海伦公式的优化忘记了，一开始写了一大坨的求根，除法，再加4个for嵌套，TLE没得说。。
赛后看了下海伦公式的优化，简直了！
题意：有N个点，找出其中四个点能够围城凸四边形的个数。
思路，根据点在三角形内和三角形外的性质来判断。
如果一个点在三角形内，改点和三条边围的三角型总面积等于三角形面积，否则大于三角形面积。
思维图
AC代码:
/*/
#include"algorithm"
#include"iostream"
#include"cstring"
#include"vector"
#include"string"
#include"cstdio"
#include"queue"
#include"cmath"
using namespace std;
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(x))
typedef long long LL;
const int MX = 1e5+5;
struct Node {
int x,y;
} nd[200];
int area(int x0,int y0,int x1,int y1,int x2,int y2 ) {
return abs(x0*y1+x1*y2+x2*y0-x2*y1-x1*y0-x0*y2);
}
bool check(int i,int j,int k,int l) {
int s[4];
s[0]=area(nd[i].x,nd[i].y,nd[j].x,nd[j].y,nd[k].x,nd[k].y);
s[1]=area(nd[i].x,nd[i].y,nd[j].x,nd[j].y,nd[l].x,nd[l].y);
s[2]=area(nd[l].x,nd[l].y,nd[j].x,nd[j].y,nd[k].x,nd[k].y);
s[3]=area(nd[i].x,nd[i].y,nd[l].x,nd[l].y,nd[k].x,nd[k].y);
sort(s,s+4);
if(s[3]!=s[2]+s[1]+s[0])return 1;
else return 0;
}
int main() {
int T,n,num;
scanf("%d",&T);
for(int qq=1; qq<=T; qq++) {
scanf("%d",&n);
num=0;
for(int i=0; i<n; i++)
scanf("%d%d",&nd[i].x,&nd[i].y);
for(int i=0; i<n; i++) {
for(int j=i+1; j<n; j++) {
for(int k=j+1; k<n; k++) {
for(int l=k+1; l<n; l++) {
// cout<<"AAA "<<num<<endl;
if(check(i,j,k,l))num++;
// cout<<"BBB "<<num<<endl;
}
}
}
}
printf("Case %d: %d
",qq,num);
}
return 0;
}