简单几何(凸包+枚举) POJ 1873 The Fortified Forest

 

题目传送门

题意：砍掉一些树，用它们做成篱笆把剩余的树围起来，问最小价值

分析：数据量不大，考虑状态压缩暴力枚举，求凸包以及计算凸包长度。虽说是水题，毕竟是final，自己状压的最大情况写错了，而且忘记特判凸包点数 <= 1的情况。

 

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/************************************************ * Author        :Running_Time * Created Time  :2015/11/3 星期二 16:10:17 * File Name     :POJ_1873.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)  {       //三态函数，减少精度问题     if (fabs (x) < EPS) return 0;     else    return x < 0 ? -1 : 1; } struct Point    {       //点的定义     double x, y, v, l;     Point () {}     Point (double x, double y, double v, double l) : x (x), y (y), v (v), l (l) {}     Point operator + (const Point &r) const {       //向量加法         return Point (x + r.x, y + r.y, 0, 0);     }     Point operator - (const Point &r) const {       //向量减法         return Point (x - r.x, y - r.y, 0, 0);     }     Point operator * (double p) const {       //向量乘以标量         return Point (x * p, y * p, 0, 0);     }     Point operator / (double p) const {       //向量除以标量         return Point (x / p, y / p, 0, 0);     }     bool operator < (const Point &r) const {       //点的坐标排序         return x < r.x || (x == r.x && y < r.y);     }     bool operator == (const Point &r) const {       //判断同一个点         return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;     } }; typedef Point Vector;       //向量的定义 Point read_point(void)   {      //点的读入     double x, y, v, l;     scanf ("%lf%lf%lf%lf", &x, &y, &v, &l);     return Point (x, y, v, l); } double dot(Vector A, Vector B)  {       //向量点积     return A.x * B.x + A.y * B.y; } double cross(Vector A, Vector B)    {       //向量叉积     return A.x * B.y - A.y * B.x; } double length(Vector A) {       //向量长度，点积     return sqrt (dot (A, A)); } Point qs[33], ps[33]; /*     点集凸包，输入点的集合，返回凸包点的集合。     如果不希望在凸包的边上有输入点，把两个 <= 改成 < */ int convex_hull(Point *ps, int n) {     sort (ps, ps+n);      //x - y排序     int k = 0;     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 i=n-2, t=k; i>=0; --i)  {         while (k > t && cross (qs[k-1] - qs[k-2], ps[i] - qs[k-1]) <= 0)  k--;         qs[k++] = ps[i];     }     return k-1; }   double cal_fence(Point *ps, int n)  {     if (n <= 1) {         return 0;     }     n = convex_hull (ps, n);     double ret = 0;     for (int i=0; i<n-1; ++i)  {         ret += length (qs[i+1] - qs[i]);     }     ret += length (qs[n-1] - qs[0]);     return ret; }   int main(void)    {     int n, cas = 0;     while (scanf ("%d", &n) == 1)   {         if (!n) break;         vector<Point> pps;         for (int i=0; i<n; ++i) {             pps.push_back (read_point ());         }         int S = 1 << n, sz = 100;         vector<int> ans, num;         int m = 0;         double sum = 0, val = 999999999, tval = 0, len = 0, ex = 0;         for (int i=0; i<S; ++i) {             num.clear ();   tval = 0;   sum = 0; m = 0;             for (int j=0; j<n; ++j) {                 if (i & (1 << j))   {       //cut                     num.push_back (j + 1);                     tval += pps[j].v;    sum += pps[j].l;                 }                 else    {                     ps[m++] = pps[j];                 }             }             len = cal_fence (ps, m);             if (sum < len)  continue;             if (val > tval || (dcmp (val - tval) == 0 && sz > num.size ())) {                 val = tval; sz = num.size ();                 ans = num;  ex = sum - len;             }         }         printf ("Forest %d\n", ++cas);         printf ("Cut these trees: ");         printf ("%d", ans[0]);         for (int i=1; i<ans.size (); ++i)   {             printf (" %d", ans[i]);         }         puts ("");         printf ("Extra wood: %.2f\n\n", ex);     }      //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";       return 0; }