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。

/************************************************
* 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;
}