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; }