简单几何(求凸包点数) POJ 1228 Grandpa's Estate

 

题目传送门

题意：判断一些点的凸包能否唯一确定

分析：如果凸包边上没有其他点，那么边想象成橡皮筋，可以往外拖动，这不是唯一确定的。还有求凸包的点数<=2的情况一定不能确定。

 

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/************************************************ * Author        :Running_Time * Created Time  :2015/11/4 星期三 10:24:45 * File Name     :POJ_1228.cpp  ************************************************/   #include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> using namespace std;   #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = acos (-1.0); int dcmp(double x)  {     if (fabs (x) < EPS) return 0;     else    return x < 0 ? -1 : 1; } struct Point    {     double x, y;     Point () {}     Point (double x, double y) : x (x), y (y) {}     Point operator - (const Point &r) const {         return Point (x - r.x, y - r.y);     }     bool operator < (const Point &r) const  {         return x < r.x || (x == r.x && y < r.y);     }     bool operator == (const Point &r) const {         return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;     } }; typedef Point Vector; Point read_point(void)  {     double x, y;    scanf ("%lf%lf", &x, &y);     return Point (x, y); } double dot(Point a, Point b)    {     return a.x * b.x + a.y * b.y; } double cross(Vector A, Vector B)    {     return A.x * B.y - A.y * B.x; } bool on_seg(Point p, Point a, Point b)    {     return dcmp (cross (a - p, b - p)) == 0 && dcmp (dot (a - p, b - p)) < 0; }   /*     凸包边上无点：<=    凸包边上有点：< */ vector<Point> convex_hull(vector<Point> ps)   {     sort (ps.begin (), ps.end ());     int n = ps.size (), k = 0;     vector<Point> qs (n * 2);     for (int i=0; i<n; ++i) {         while (k > 1 && cross (qs[k-1] - qs[k-2], ps[i] - qs[k-1]) <= 0)    k--;         qs[k++] = ps[i];     }     for (int t=k, i=n-2; i>=0; --i) {         while (k > t && cross (qs[k-1] - qs[k-2], ps[i] - qs[k-1]) <= 0)     k--;         qs[k++] = ps[i];     }     qs.resize (k - 1);     return qs; }   int main(void)    {     int T;  scanf ("%d", &T);     while (T--) {         int n;  scanf ("%d", &n);         vector<Point> ps;         for (int i=0; i<n; ++i) ps.push_back (read_point ());         if (n == 1) {             puts ("NO");    continue;         }         vector<Point> qs = convex_hull (ps);         if (qs.size () == n || qs.size () <= 2)    {             puts ("NO");    continue;         }         qs.push_back (qs[0]);         int m = qs.size ();         bool flag = false;         for (int i=0; i<m-1; ++i)   {             flag = false;             for (int j=0; j<ps.size (); ++j)    {                 if (ps[j] == qs[i] || ps[j] == qs[i+1]) continue;                 if (on_seg (ps[j], qs[i], qs[i+1])) {                     flag = true;    break;                 }             }             if (!flag)  break;         }         if (flag)   puts ("YES");         else    puts ("NO");     }      //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";       return 0; }