FZU 2148 Moon Game

URL: http://acm.fzu.edu.cn/problem.php? Pid = 1, 2148

PS: When we enumerate the four points, the result is a convex quadrilateral. The method is to solve a convex packet.

#include 
 
  #include 
  
   #include 
   
    #include #include 
    
     const int maxn  = 40;using namespace std;//Accepted2148  GNU C++  656 ms228KB2154Bachiberxstruct point {    int x, y;    point(int a=0, int b=0):x(a), y(b){}    int operator^(const point &op) const {        return x*op.y - y*op.x;    }};point p[maxn];int n;point tmp[8];point p0;int sqr(int x) {    return x*x;}int dis2(point p0, point p1) {    return sqr(p0.x-p1.x)+sqr(p0.y-p1.y);}point operator - (point A, point B) {    return point(A.x-B.x, A.y-B.y);}int mul(point p0, point p1, point p2) {    return (p1-p0)^(p2-p0);}bool cmp(point a, point b) {    if(mul(p0, a, b)>0 || (mul(p0, a, b)==0 && dis2(p0, a)
     
      1 && mul(q[newn-1], q[newn-2], tmp[i])>=0)            --newn;        q[newn++] = tmp[i];    }    if(newn==4) return true;    else return false;}int main(){    int T;    int cnt = 0;    scanf("%d", &T);    for(int t = 1; t <= T; t++) {        scanf("%d", &n);        for(int i = 0; i < n; i++) {            scanf("%d%d", &p[i].x, &p[i].y);        }        cnt = 0;        if(n<=3) {            printf("Case %d: %d\n", t, cnt);            continue;        }        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 q = k+1; q < n; q++) {                        bool res = work(p[i], p[j], p[k], p[q]);                        if(res) cnt++;                    }                }            }        }        printf("Case %d: %d\n", t, cnt);    }    return 0;}