FZU

Problem 2148 Moon Game

Accept: 1040    Submit: 3205
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

题意：给若干个点，求这些点能组成凸四边形的个数

解决：枚举每种情况，用面积判断：如果是凹四边形，凹进去的那个点就在另外三个点组成的三角形里面，这个点与另外两个点组成的三个三角形面积和等于最外面那个三角形的面积。

细节：坐标求三角形面积公式：S=(1/2)*(x1y2+x2y3+x3y1-x1y3-x2y1-x3y2)

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <set>#include <stack>#include <queue>#include <cmath>#include <map>#define fin freopen("in.txt", "r", stdin)#define line printf("-----------------\n")using namespace std;const int maxn = 1000000;int t, n, x, y;struct Point{    int x, y;    Point(int xx = 0, int yy = 0){       this-> x = xx; this->y = yy;    }}point[35];double area(int a, int b, int c){    //S=(1/2)*(x1y2+x2y3+x3y1-x1y3-x2y1-x3y2)    int ar = point[a].x*point[b].y + point[b].x*point[c].y + point[c].x*point[a].y          - point[a].x*point[c].y - point[b].x*point[a].y - point[c].x*point[b].y;    return abs(ar) * 0.5;}bool ok(int a, int b, int c, int d){    double aa = area(a, b, c), bb = area(a, b, d), cc = area(a, c, d), dd = area(b, c, d);    if(2*max(max(aa, bb), max(cc,dd)) == aa+bb+cc+dd) return false;    return true;}int main(){    cin >> t;    for(int cases = 1; cases <= t; cases++){        cin >> n;        int num = 0;        for(int i = 0; i < n; i++){            cin >> x >> y;            point[i] = Point(x,y);        }        for(int i = 0; i < n-3; i++)        for(int j = i+1; j < n-2; j++)        for(int k = j+1; k < n-1; k++)        for(int o = k+1; o < n; o++)            if(ok(i,j,k,o)) num++;        printf("Case %d: %d\n", cases, num);    }}