Poj 1873 The Fortified Forest

题目：

The Fortified Forest

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 6421		Accepted: 1811

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.

Sample Input

Sample Output

Forest 1
Cut these trees: 2 4 5 
Extra wood: 3.16

Forest 2
Cut these trees: 2 
Extra wood: 15.00

思路：这题思路很简单枚举+凸包就好了，不过我还是wa了好久，，，凸包卷积法的模板挂了

　　我的模板有毒！

  1 /* 二维几何  */
  2 /* 需要包含的头文件 */
  3 #include<cstdio>
  4 #include <cstring>
  5 #include <cmath >
  6 #include <iostream>
  7 #include <algorithm>
  8 #include <vector>
  9 using namespace std;
 10 /** 常用的常量定义 **/
 11 const double INF  = 1e200;
 12 const double eps  = 1e-8;
 13 const double PI  = acos(-1.0);
 14 const int Max = 1e6;
 15 /** 基本几何结构 **/
 16 struct Point
 17 {
 18     double x,y;
 19     Point(double a=0, double b=0){x=a,y=b;}
 20     bool operator<(const Point &ta)const
 21     {
 22         if(x==ta.x)     return y<ta.y;
 23         return x<ta.x;
 24     }
 25     friend Point operator+(const Point &ta,const Point &tb)
 26     {
 27         return Point(ta.x+tb.x,ta.y+tb.y);
 28     }
 29     friend Point operator-(const Point &ta,const Point &tb)
 30     {
 31         return Point(ta.x-tb.x,ta.y-tb.y);
 32     }
 33 };
 34 struct LineSeg      ///线段，重载了/作为叉乘运算符，*作为点乘运算符
 35 {
 36     Point s,e;
 37     LineSeg(){s=Point(0,0),e=Point(0,0);}
 38     LineSeg(Point a, Point b){s=a,e=b;}
 39     double lenth(void)
 40     {
 41         return sqrt((s.x-e.x)*(s.x-e.x)+(s.y-e.y)*(s.y-e.y));
 42     }
 43     friend double operator*(const LineSeg &ta,const LineSeg &tb)
 44     {
 45         return (ta.e.x-ta.s.x)*(tb.e.x-tb.s.x)+(ta.e.y-ta.s.y)*(tb.e.y-tb.s.y);
 46     }
 47     friend double operator/(const LineSeg &ta,const LineSeg &tb)
 48     {
 49         return (ta.e.x-ta.s.x)*(tb.e.y-tb.s.y)-(ta.e.y-ta.s.y)*(tb.e.x-tb.s.x);
 50     }
 51     LineSeg operator=(const LineSeg &ta)
 52     {
 53         s=ta.s,e=ta.e;
 54         return *this;
 55     }
 56 };
 57 
 58 int sgn(double ta,double tb);
 59 double getdis(const Point &ta,const Point &tb);
 60 double graham(Point tb[],double len);
 61 
 62 int n,miv,v[20],cut[2][20];
 63 double ex,use[20];
 64 Point pt[20],tb[20];
 65 vector<Point>va;
 66 int main(void)
 67 {
 68     int k=0;
 69     while(1==scanf("%d",&n)&&n)
 70     {
 71         miv=1e9,ex=0;
 72         for(int i=0;i<n;i++)
 73             scanf("%lf%lf%d%lf",&pt[i].x,&pt[i].y,&v[i],&use[i]);
 74         for(int i=1,sz=(1<<n)-1;i<sz;i++)
 75         {
 76             va.clear();
 77             int tv=0;
 78             double len=0,lf=0;
 79             cut[1][0]=0;
 80             for(int j=0;j<n;j++)
 81             if(i&(1<<j))
 82                 cut[1][++cut[1][0]]=j,len+=use[j],tv+=v[j];
 83             else
 84                 va.push_back(pt[j]);
 85             if(cut[1][0]==3)
 86             {
 87                 134;
 88             }
 89             lf=graham(tb,len);
 90             if(sgn(lf,0)<0)
 91                 continue;
 92             if(miv>tv ||(tv==miv&&cut[1][0]<cut[0][0]))
 93             {
 94                 for(int j=0,ls=cut[1][0];j<=ls;j++)
 95                     cut[0][j]=cut[1][j];
 96                 miv=tv,ex=lf;
 97             }
 98 
 99         }
100         printf("Forest %d
Cut these trees:",++k);
101         for(int i=1;i<=cut[0][0];i++)
102             printf(" %d",1+cut[0][i]);
103         printf("
Extra wood: %.2f

",ex);
104     }
105     return 0;
106 }
107 
108 
109 
110 
111 /*******判断ta与tb的大小关系*******/
112 int sgn(double ta,double tb)
113 {
114     if(fabs(ta-tb)<eps)return 0;
115     if(ta<tb)   return -1;
116     return 1;
117 }
118 /*********求两点的距离*************/
119 double getdis(const Point &ta,const Point &tb)
120 {
121     return sqrt((ta.x-tb.x)*(ta.x-tb.x)+(ta.y-tb.y)*(ta.y-tb.y));
122 }
123 
124 bool cmp(const Point &ta,const Point &tb)/// 选取与最后一条确定边夹角最小的点，即余弦值最大者
125 {
126     double tmp=LineSeg(va[0],ta)/LineSeg(va[0],tb);
127     if(sgn(tmp,0)==0)
128         return getdis(va[0],ta)<getdis(va[0],tb);
129     else if(tmp>0)
130         return 1;
131     return 0;
132 }
133 double graham(Point tb[],double len)
134 {
135     if(va.size()==1)  return   len;
136     else if(va.size()==2) return len-2.0*getdis(va[0],va[1]);
137     int cur=0,top=2;
138     for(int i=1;i<va.size();i++)
139         if(sgn(va[cur].y,va[i].y)>0 || (sgn(va[cur].y,va[i].y)==0 && sgn(va[cur].x,va[i].x)>0))
140             cur=i;
141     swap(va[cur],va[0]);
142     sort(va.begin()+1,va.end(),cmp);
143     tb[0]=va[0],tb[1]=va[1],tb[2]=va[2];
144     for(int i=3;i<va.size();i++)
145     {
146         while(sgn(LineSeg(tb[top-1],tb[top])/LineSeg(tb[top-1],va[i]),0)<0)
147             top--;
148         tb[++top]=va[i];
149     }
150     double tmp=0;
151     tb[++top]=tb[0];
152     for(int i=0;i<top;i++)
153         tmp+=getdis(tb[i],tb[i+1]);
154     return len-tmp;
155 }