The Fortified Forest
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 6198 | Accepted: 1744 |
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.
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).
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.
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
6 0 0 8 3 1 4 3 2 2 1 7 1 4 1 2 3 3 5 4 6 2 3 9 8 3 3 0 10 2 5 5 20 25 7 -3 30 32 0
Sample Output
Forest 1 Cut these trees: 2 4 5 Extra wood: 3.16 Forest 2 Cut these trees: 2 Extra wood: 15.00
/*
poj 1873 凸包+枚举
给你n棵树,已知树的树的长度以及他们的价值。要砍掉一些树来给剩下的数围一个篱笆
要求剩下的数价值尽可能大,如果价值相同则希望剩下的树尽可能多
因为最多15棵,枚举需要砍掉的树,然后通过凸包判断是否能围成,记录一下即可
hhh-2016-05-07 21:27:41
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <cmath>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson (i<<1)
#define rson ((i<<1)|1)
typedef long long ll;
using namespace std;
const int maxn = 20;
double PI = 3.1415926;
double eps = 1e-8;
int sgn(double x)
{
if(fabs(x) < eps) return 0;
if(x < 0)
return -1;
else
return 1;
}
struct Point
{
double x,y;
Point() {}
Point(double _x,double _y)
{
x = _x,y = _y;
}
Point operator -(const Point &b)const
{
return Point(x-b.x,y-b.y);
}
double operator ^(const Point &b)const
{
return x*b.y-y*b.x;
}
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
};
struct Line
{
Point s,t;
Line() {}
Line(Point _s,Point _t)
{
s = _s;
t = _t;
}
pair<int,Point> operator &(const Line&b)const
{
Point res = s;
if( sgn((s-t) ^ (b.s-b.t)) == 0) //通过叉积判断
{
if( sgn((s-b.t) ^ (b.s-b.t)) == 0)
return make_pair(0,res);
else
return make_pair(1,res);
}
double ta = ((s-b.s)^(b.s-b.t))/((s-t)^(b.s-b.t));
res.x += (t.x-s.x)*ta;
res.y += (t.y-s.y)*ta;
return make_pair(2,res);
}
};
Point lis[maxn];
int Stack[maxn],top;
double dist(Point a,Point b)
{
return sqrt((a-b)*(a-b));
}
Point ta[20];
bool cmp(Point a,Point b)
{
double t = (a-ta[0])^(b-ta[0]);
if(sgn(t) == 0)
{
return dist(a,ta[0]) <= dist(b,ta[0]);
}
if(sgn(t) < 0)
return false;
else
return true;
}
int tot;
double Graham(int n)
{
Point p;
if(n == 1 || n == 0)
{
return 0;
}
if(n == 2)
{
return dist(ta[0],ta[1])*2;
}
int k = 0;
p = ta[0];
for(int i = 1; i < n; i++)
{
if(p.y > ta[i].y || (p.y == ta[i].y && p.x > ta[i].x))
p = ta[i],k = i;
}
swap(ta[0],ta[k]);
sort(ta+1,ta+n,cmp);
Stack[0] = 0;
Stack[1] = 1;
top = 2;
for(int i = 2; i < n; i++)
{
while(top > 1 && sgn((ta[Stack[top-1]]-ta[Stack[top-2]])
^ (ta[i]-ta[Stack[top-2]])) <= 0)
top --;
Stack[top++] = i;
}
double len = 0;
for(int i = 0; i < top; i++)
{
if(i == top - 1)
len += dist(ta[Stack[i]],ta[Stack[0]]);
else
len += dist(ta[Stack[i]],ta[Stack[i+1]]);
}
return len;
}
int val[maxn];
double lent[maxn];
int main()
{
//freopen("in.txt","r",stdin);
int n;
int cas = 1;
while(scanf("%d",&n) && n)
{
if(cas != 1)
printf("
");
for(int i = 0; i < n; i++)
{
scanf("%lf%lf%d%lf",&lis[i].x,&lis[i].y,&val[i],&lent[i]);
}
int ansV = 0x7fffffff,ansN = 0x7fffffff,ansX = 0;
double ansL = 0;
for(int i = 0; i < (1<<n); i++)
{
tot = 0;
double lans = 0;
int vans = 0;
for(int j = 0; j < n; j++)
{
if(i & (1 << j))
{
vans += val[j];
lans += lent[j];
}
else
{
ta[tot++] = lis[j];
}
}
if(vans > ansV)
continue;
double t = Graham(tot);
if(lans >= t)
{
if(vans < ansV || (n-tot < ansN && vans == ansV))
{
ansV = vans;
ansL = lans-t;
ansX = i;
ansN = n-tot;
}
}
}
printf("Forest %d
",cas++);
printf("Cut these trees:");
for(int i = 0; i < n; i++)
{
if(ansX&(1 << i))
printf(" %d",i+1);
}
printf("
");
printf("Extra wood: %.2f
",ansL);
}
return 0;
}