poj1873 The Fortified Forest【枚举+凸包】

来源：互联网发布：关于绘画的软件编辑：程序博客网时间：2024/05/29 10:57

题目链接：http://poj.org/problem?id=1873
题意：给你n棵树，让你从这n棵树中，看k棵树，把剩下的数围起来，砍每棵树的价值和每棵树的长度都告诉你，问你能否话最小的价值把剩下的书围起来，如果价值相同，则输出看的树最少的结果。最后需要输出的是，砍了哪几颗树，还剩下都少长度的木材
解析：由于n才15，于是可以考虑枚举砍的情况的所有子集，然后把剩下的树做凸包，然后不断更新答案即可

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>using namespace std;const int maxn = 1e5+100;const int inf = 0x7fffffff;const double eps = 1e-10;const double pi = acos(-1.0);struct point{    double x,y;    double v,l;    point() {}    point(double _x,double _y,double _v,double _l)    {        x = _x;        y = _y;        v = _v;        l = _l;    }    bool operator < (const point &b) const    {        if(y==b.y)            return x<b.x;        return y<b.y;    }    bool operator == (const point &b)const    {        return x==b.x && y==b.y;    }}t1[maxn],a[maxn],ans[maxn];double x_mul(point p0,point p1,point p2){    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}double dis(point p1,point p2){    return sqrt((p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));}bool cmp(point t1,point t2){    double tmp = x_mul(a[0],t1,t2);    if(tmp==0)        return dis(a[0],t1)<dis(a[0],t2);    else        return tmp>0;}double graham(int n){    if(n==0 || n==1)        return 0;    if (n == 2)        return dis(a[0], a[1])*2;    sort(a,a+n);    sort(a+1,a+n,cmp);    ans[0] = a[0];    ans[1] = a[1];    ans[2] = a[2];    int top = 2;    for(int i=3;i<n;i++)    {        while(top>=2&&x_mul(ans[top-1],ans[top],a[i])<0)            top--;        ans[++top] = a[i];    }    double res = 0;    for(int i=0;i<top;i++)        res += dis(ans[i],ans[i+1]);    res += dis(ans[top],ans[0]);    return res;}int main(void){    int n;    int case_t = 0;    while(~scanf("%d",&n)&&n)    {        if(case_t)            puts("");        for(int i=0;i<n;i++)            scanf("%lf %lf %lf %lf",&t1[i].x,&t1[i].y,&t1[i].v,&t1[i].l);        double ans1 = inf;        int ans = 0;        double ans_len,cur = 0;        for(int i=0;i<(1<<n);i++)        {            int cnt = 0;            double v = 0,len = 0;            for(int j=0;j<n;j++)            {                if(i&(1<<j))                {                    v += t1[j].v;                    len += t1[j].l;                }                else                    a[cnt++] = t1[j];            }            if(v>ans1)                continue;            double tmp = graham(cnt);            if(len>=tmp)            {                if(v<ans1 || (v==ans1 && cur>cnt))                {                    ans1 = v;                    ans = i;                    cur = cnt;                    ans_len = len-tmp;                }            }        }        printf("Forest %d\n",++case_t);        printf("Cut these trees:");        for(int i=0;i<n;i++)        {            if(ans&(1<<i))                printf(" %d",i+1);        }        printf("\nExtra wood: %.2f\n",ans_len);    }    return 0;}

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