POJ 1873 The Fortified Forest

POJ_1873

    由于树木比较少，可以枚举砍树的决策，然后计算砍掉的树木的长度是否比剩下的树木的凸包的周长要长，每次计算之后按题意更新最优解即可。

    在枚举的时候还可以应用一个剪枝，就是如果当前砍树损失的价值比记录的最小价值要大的话，就可以直接枚举下一种情况了，而不必再去花时间计算凸包和凸包的周长了，这样可以节省很多时间。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#define MAXD 20
#define zero 1e-8
#define INF 0x3f3f3f3f
struct point
{
    int x, y;
}p[MAXD], res[MAXD], q[MAXD];
int st, N, X, P, num, minv, v[MAXD];
double w[MAXD], ans;
int cmp(const void *_p, const void *_q)
{
    point *p = (point *)_p, *q = (point *)_q;
    if(p->y == q->y)
        return p->x - q->x;
    return p->y - q->y;
}
int dcmp(double x)
{
    return fabs(x) < 0 ? 0 : (x < 0 ? -1 : 1);
}
double sqr(double x)
{
    return x * x;
}
int det(int x1, int y1, int x2, int y2)
{
    return x1 * y2 - x2 * y1;
}
void init()
{
    int i, j, k;
    for(i = 0; i < N; i ++)
        scanf("%d%d%d%lf", &p[i].x, &p[i].y, &v[i], &w[i]);
}
int del(int top, int i)
{
    if(det(res[top].x - res[top - 1].x, res[top].y - res[top - 1].y, q[i].x - res[top].x, q[i].y - res[top].y) <= 0)
        return 1;
    return 0;
}
int graham()
{
    int i, j, k, top = 1, mint;
    qsort(q, X, sizeof(q[0]), cmp);
    res[0] = q[0], res[1] = q[1];
    for(i = 2; i < X; i ++)
    {
        while(top && del(top, i))
            -- top;
        res[++ top] = q[i];
    }
    mint = top;
    res[++ top] = q[X - 2];
    for(i = X - 3; i >= 0; i --)
    {
        while(top != mint && del(top, i))
            -- top;
        res[++ top] = q[i];
    }
    return top;
}
double calculate()
{
    int i, j, k;
    double t;
    if(X == 1)
        t = 0;
    else if(X == 2)
        t = 2 * sqrt(sqr(q[0].x - q[1].x) + sqr(q[0].y - q[1].y));
    else
    {
        P = graham();
        t = 0;
        for(i = 0; i < P; i ++)
            t += sqrt(sqr(res[i].x - res[i + 1].x) + sqr(res[i].y - res[i + 1].y));
    }
    return t;
}
void solve()
{
    int i, j, k, cnt, value;
    double t, s;
    minv = num = INF;
    for(k = 1; k < (1 << N); k ++)
    {
        s = cnt = value = X = 0;
        for(j = 0; j < N; j ++)
        {
            if((1 << j) & k)
            {
                ++ cnt;
                value += v[j];
                s += w[j];
            }
            else
                q[X ++] = p[j];
        }
        if(value > minv)
            continue;
        t = calculate();
        if(dcmp(s - t) < 0)
            continue;
        if(value < minv || (value == minv && cnt < num))
        {
            minv = value;
            num = cnt;
            st = k;
            ans = s - t;
        }
    }
    printf("Cut these trees:");
    for(i = 0; i < N; i ++)
        if((1 << i) & st)
            printf(" %d", i + 1);
    printf("\n");
    printf("Extra wood: %.2lf\n", ans);
}
int main()
{
    int t = 0;
    for(;;)
    {
        scanf("%d", &N);
        if(!N)
            break;
        init();
        if(t ++)
            printf("\n");
        printf("Forest %d\n", t);
        solve();
    }
    return 0;
}