POJ 1873 The Fortified Forest（枚举+凸包）

Description

Once upon a time, in a faraway land, there lived a king. This king owned a small collection of rare and valuable trees, which had been gathered by his ancestors on their travels. To protect his trees from thieves, the king ordered that a high fence be built around them. His wizard was put in charge of the operation. 
Alas, the wizard quickly noticed that the only suitable material available to build the fence was the wood from the trees themselves. In other words, it was necessary to cut down some trees in order to build a fence around the remaining trees. Of course, to prevent his head from being chopped off, the wizard wanted to minimize the value of the trees that had to be cut. The wizard went to his tower and stayed there until he had found the best possible solution to the problem. The fence was then built and everyone lived happily ever after.

You are to write a program that solves the problem the wizard faced.

Input

The input contains several test cases, each of which describes a hypothetical forest. Each test case begins with a line containing a single integer n, 2 <= n <= 15, the number of trees in the forest. The trees are identified by consecutive integers 1 to n. Each of the subsequent n lines contains 4 integers xi, yi, vi, li that describe a single tree. (xi, yi) is the position of the tree in the plane, vi is its value, and li is the length of fence that can be built using the wood of the tree. vi and li are between 0 and 10,000. 
The input ends with an empty test case (n = 0). 

Output

For each test case, compute a subset of the trees such that, using the wood from that subset, the remaining trees can be enclosed in a single fence. Find the subset with minimum value. If more than one such minimum-value subset exists, choose one with the smallest number of trees. For simplicity, regard the trees as having zero diameter. 
Display, as shown below, the test case numbers (1, 2, ...), the identity of each tree to be cut, and the length of the excess fencing (accurate to two fractional digits).

Display a blank line between test cases.

 

题目大意：有n棵树，每棵树有坐标(x,y)，价值v，长度l，问如何砍能砍掉最小价值为的树（价值相同则砍最少的树），能把其他树都围起来
思路：枚举所有砍树的方案（我用的递归，用二进制的方法理论上来说也可以），算一下能不能围起剩下的树（如果价值比当前答案要大就不用算了）。至于怎么围起剩下的树，一个点的明显是需要0长度，两个点就需要这两个点的距离*2，三个点或以上就要用到求凸包的方法（反正我的凸包是不能算三个点以下的）

PS：输出最好复制啊，我好像就是因为forest打错了WA了好几次啊……

 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 using namespace std;

 const double EPS = 1e-;

 inline int sgn(const double &x) {
     if(fabs(x) < EPS) return ;
     return x >  ?  : -;
 }

 struct Point {
     double x, y;
     int v, l;
 };

 inline bool Cross(Point &sp, Point &ep, Point &op) {
     return (sp.x - op.x) * (ep.y - op.y) - (ep.x - op.x) * (sp.y - op.y) >= ;
 }

 inline double dist(Point &a, Point &b) {
     return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
 }

 inline bool cmp(const Point &a, const Point &b) {
     if(a.y == b.y) return a.x < b.x;
     return a.y < b.y;
 }

 const int MAXN = ;
 int stk[MAXN];
 bool cut[MAXN], ans[MAXN];
 Point p[MAXN], a[MAXN];
 int n, top;
 double answood;

 double Graham_scan(int n) {
     sort(p, p + n, cmp);
     top = ;
     stk[] = ; stk[] = ;
     for(int i = ; i < n; ++i) {
         while(top && Cross(p[i], p[stk[top]], p[stk[top - ]])) --top;
         stk[++top] = i;
     }
     int len = top;
     stk[++top] = n - ;
     for(int i = n - ; i >= ; --i) {
         while(top != len && Cross(p[i], p[stk[top]], p[stk[top - ]])) --top;
         stk[++top] = i;
     }
     double sum = ;
     stk[++top] = stk[];
     for(int i = ; i < top; ++i)
         sum += dist(p[stk[i]], p[stk[i+]]);
     return sum;
 }

 int minval, mincut, sumval, sumlen;
 double uselen;

 void setans(int cutcnt) {
     for(int i = ; i <= n; ++i) ans[i] = cut[i];
     minval = sumval;
     mincut = cutcnt;
     answood = sumlen - uselen;
 }

 void dfs(int dep, int cutcnt) {
     if(dep == n + ) {
         if(n == cutcnt) return ;
         sumval = sumlen = ;
         for(int i = ; i <= n; ++i) {
             if(!cut[i]) continue;
             sumval += a[i].v;
             sumlen += a[i].l;
         }
         if(sumval > minval) return ;
         if(sumval == minval && cutcnt >= mincut) return ;
         if(n - cutcnt == ) {
             uselen = ;
             setans(cutcnt);
         }
         else if(n - cutcnt == ) {
             int i1 = , i2 = ;
             for(int i = ; i <= n; ++i) {
                 if(cut[i]) continue;
                 if(!i1) i1 = i;
                 else i2 = i;
             }
             uselen =  * dist(a[i1], a[i2]);
             if(uselen <= sumlen) setans(cutcnt);
         }
         else {
             int pcnt = ;
             for(int i = ; i <= n; ++i) {
                 if(cut[i]) continue;
                 p[pcnt++] = a[i];
             }
             uselen = Graham_scan(pcnt);
             if(sgn(uselen - sumlen) <= ) setans(cutcnt);
         }
         return ;
     }
     cut[dep] = false;
     dfs(dep + , cutcnt);
     cut[dep] = true;
     dfs(dep + , cutcnt + );
 }

 int main() {
     int ca = ;
     while(scanf("%d", &n) != EOF && n) {
         for(int i = ; i <= n; ++i) {
             scanf("%lf%lf%d%d", &a[i].x, &a[i].y, &a[i].v, &a[i].l);
         }
         mincut = MAXN;
         minval = 0x7fffffff;
         dfs(, );
         if(ca != ) printf("\n");
         printf("Forest %d\n", ca++);
         printf("Cut these trees:");
         for(int i = ; i <= n; ++i) if(ans[i]) printf(" %d", i);
         printf("\nExtra wood: %.2f\n", answood);
     }
 }