题意：原来一个凸多边形删去一些点后剩n个点，问这个n个点能否确定原来的凸包(1 <= 测试组数t <= 10，1 <= n <= 1000)。
题目链接：http://poj.org/problem?id=1228
——>>初看这题，好别扭，不知道要做什么。。。
其实，是这样的：先求凸包，然后看凸包每一条边所在直线上有多少个点，至少需要3个。
假设一条边的所在直线只有2个点，那么可适当地在这两个点中间加一个或者几个点，使新图形仍是凸包，这时候就不能确定原来的凸包了。
假设一条边的所在直线上有3个以上的点，如果在其中两点间扩展一个点，所形成的图形是凹的，所以不能扩展，即边就确定了。
#include <cstdio>
#include <cmath>
#include <algorithm>using namespace std;const int maxn = 1000 + 10;
const double eps = 1e-10;int dcmp(double x){if(fabs(x) < eps) return 0;else return x < 0 ? -1 : 1;
}struct Point{double x;double y;Point(double x = 0, double y = 0):x(x), y(y){}bool operator < (const Point& e) const{return x < e.x || (dcmp(x - e.x) == 0 && y < e.y);}
}p[maxn], q[maxn];typedef Point Vector;Vector operator + (Point A, Point B){return Vector(A.x + B.x, A.y + B.y);
}Vector operator - (Point A, Point B){return Vector(A.x - B.x, A.y - B.y);
}Vector operator * (Point A, double p){return Vector(A.x * p, A.y * p);
}Vector operator / (Point A, double p){return Vector(A.x / p, A.y / p);
}double Cross(Vector A, Vector B){return A.x * B.y - B.x * A.y;
}int ConvexHull(Point *p, int n, Point* ch){ //求凸包sort(p, p + n);int m = 0;for(int i = 0; i < n; i++){while(m > 1 && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) < 0) m--;ch[m++] = p[i];}int k = m;for(int i = n-2; i >= 0; i--){while(m > k && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) < 0) m--;ch[m++] = p[i];}if(n > 1) m--;return m;
}double ConvexPolygonArea(Point *p, int n){double area = 0;for(int i = 1; i < n-1; i++) area += Cross(p[i]-p[0], p[i+1]-p[0]);return area / 2;
}int main()
{int t, n;scanf("%d", &t);while(t--){bool ok = 1;scanf("%d", &n);for(int i = 0; i < n; i++) scanf("%lf%lf", &p[i].x, &p[i].y);if(n < 3) ok = 0;else{int m = ConvexHull(p, n, q);if(m < 6) ok = 0;if(ok) for(int i = 2; i < m; i++){if(dcmp(Cross(q[i] - q[i-1], q[i] - q[i-2])) != 0){ok = 0;break;}while(dcmp(Cross(q[i] - q[i-1], q[i] - q[i-2])) == 0) i++;}if(dcmp(ConvexPolygonArea(q, m)) == 0) ok = 0;}if(ok) puts("YES");else puts("NO");}return 0;
}