POJ 1873 The Fortified Forest（凸包+枚举 World Finals 1999啊）

题目链接：http://poj.org/problem?id=1873

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

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

Source

World Finals 1999

题意：

有n棵树（n<=15）, 每棵树的价值为 v , 砍掉这个树能得到长度为 l 的木板。

现在要砍掉其中的一些树，用来造木板把剩余的所有树围起来,(每颗树只有砍与不砍两种状态)，

问要造这种木板最少需要消耗多少的价值(砍掉的树的 v 的和), 

输出需要砍掉哪些树和剩余的木板长度（砍掉的树形成的木板并且包围未砍掉的树没有用完的木板长度）。

PS:

n比较小，我们可以枚举砍掉的树！

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
using namespace std;
#define INF 0x3f3f3f3f
const double eps = 1e-8;
const double PI = acos(-1.0);
const int MAXN = 1010;
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}
struct Point
{
    int x, y;
    int v, l;
    Point() {}
    Point(int _x, int _y)
    {
        x = _x;
        y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x,y - b.y);
    }
//叉积
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
//点积
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
};
Point list[MAXN], tr[MAXN];
int Stack[MAXN], top;
//*两点间距离
double dist(Point a,Point b)
{
    return sqrt((a-b)*(a-b));
}
//相对于list[0]的极角排序
bool _cmp(Point p1,Point p2)
{
    double tmp = (p1-list[0])^(p2-list[0]);
    if(sgn(tmp) > 0)return true;
    else if(sgn(tmp) == 0 && sgn(dist(p1,list[0]) - dist(p2,list[0])) <= 0)
        return true;
    else return false;
}
void Graham(int n)
{
    Point p0;
    int k = 0;
    p0 = list[0];
//找最下边的一个点
    for(int i = 1; i < n; i++)
    {
        if( (p0.y > list[i].y) || (p0.y == list[i].y && p0.x > list[i].x) )
        {
            p0 = list[i];
            k = i;
        }
    }
    swap(list[k],list[0]);
    sort(list+1,list+n,_cmp);
    if(n == 1)
    {
        top = 1;
        Stack[0] = 0;
        return;
    }
    if(n == 2)
    {
        top = 2;
        Stack[0] = 0;
        Stack[1] = 1;
        return ;
    }
    Stack[0] = 0;
    Stack[1] = 1;
    top = 2;
    for(int i = 2; i < n; i++)
    {
        while(top > 1 &&
                sgn((list[Stack[top-1]]-list[Stack[top-2]])^(list[i]-list[Stack[top-2]])) <=
                0)
            top--;
        Stack[top++] = i;
    }
}
int main()
{
    int n;
    int cas = 0;
    int ans[20];
    while(scanf("%d",&n) && n)
    {
        int x, y, v, l;
        int pnum;//最少砍掉的数量
        double rest = 0;//剩余的长度
        int min_tree = n+1;
        int min_val = INF;
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d%d%d",&tr[i].x,&tr[i].y,&tr[i].v,&tr[i].l);
        }
        for(int i = 0; i < (1<<n); i++)
        {
            double len, val;
            int num;
            len = num = val = 0;
            int k = 0;
            int vis[27];
            memset(list,0,sizeof(list));
            for(int j = 0; j < n; j++)
            {
                vis[j] = 0;
                if(i & (1<<j))//砍掉
                {
                    val+=tr[j].v;
                    len+=tr[j].l;
                    num++;
                    vis[j] = 1;
                }
                else
                {
                    list[k++] = Point(tr[j].x, tr[j].y);
                }
            }
            if(val > min_val || (val==min_val && min_tree < num))
            {
                continue;
            }
            if(len == 0 && k != 1)
            {
                continue;
            }
            Graham(k);
            double res = 0;//凸包周长
            for(int h = 0; h < top-1; h++)
            {
                res+=dist(list[Stack[h]],list[Stack[h+1]]);
            }
            res+=dist(list[Stack[0]],list[Stack[top-1]]);

            //printf("len::%.2lf >>res:%.2lf, ans::%.2lf\n",len,res,len-res);
            if(res <= len)//能包围
            {
                min_val = val;
                min_tree = num;
                rest = len - res;
                pnum = 0;
                for(int f = 0; f < n; f++)
                {
                    if(vis[f])
                    {
                        ans[pnum++] = f;
                    }
                }
            }
        }
        printf("Forest %d\n",++cas);
        printf("Cut these trees:");
        for(int i = 0; i < pnum; i++)
        {
            printf(" %d",ans[i]+1);
        }
        printf("\nExtra wood: %.2f\n\n", rest);
    }
}