POJ 1873 凸包+枚举

题意：给你n棵树，每棵树都有它的坐标，价值，长度。现在要砍掉一些树，然后用这些树产生的木材（就是长度和）能够把其他的树围起来（剩下的树的凸包的周长），找到价值最小的解，如果价值相同，就选砍掉较少的树的解

思路：就是枚举+求凸包周长，需要有个剪纸就是如果当前的价值比当前最小价值大，那么这个情况肯定是不行的，不然会T

水的第一道final题= =

代码：

#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<cmath>#include<queue>#include<list>using namespace std;const double INF = 1e200;const double EP = 1e-10;const int maxn = 16;const double PI = acos(-1);struct POINT{///点 定义    double x;    double y;    int v,l;    POINT(double a = 0,double b = 0,int vv = 0,int ll = 0){x = a;y = b;v = vv;l = ll;}};struct SEGMENT{///line segment///线段 定义    POINT s;    POINT e;    SEGMENT(POINT a,POINT b){s = a;e = b;}    SEGMENT(){}};struct LINE{///ax + by + c = 0&&a >= 0///一般式    double a;    double b;    double c;    LINE(double da,double db,double dc){a = da;b = db;c = dc;}    LINE(double x1,double y1,double x2,double y2){///根据两个点求出一般式        a = y1 - y2;b = x2 - x1;c = x1*y2 - x2*y1;        if(a < 0){a*=-1;b*=-1;c*=-1;}    }};double multiply(POINT sp,POINT ep,POINT op){///向量op->sp X op->ep的叉乘，小于0：ep在op->sp顺时针方向//大于0：ep在op->sp逆时针方向//等于0：三点共线    return ((sp.x - op.x)*(ep.y - op.y) - (ep.x - op.x)*(sp.y - op.y));}double dotmultiply(POINT p1,POINT p2,POINT p0){    return ((p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y));}bool online(SEGMENT l,POINT p){///判断点是否在线段上    return ((multiply(l.e,p,l.s) == 0)&&(((p.x-l.s.x)*(p.x-l.e.x))<= 0)&&(((p.y-l.s.y)*(p.y-l.e.y)) <= 0));}bool intersect(SEGMENT u,SEGMENT v){///两线段相交（包括端点），返回true    return ((max(u.s.x,u.e.x) >= min(v.s.x,v.e.x))&&            (max(v.s.x,v.e.x) >= min(u.s.x,u.e.x))&&            (max(u.s.y,u.e.y) >= min(v.s.y,v.e.y))&&            (max(v.s.y,v.e.y) >= min(u.s.y,u.e.y))&&            (multiply(v.s,u.e,u.s)*multiply(u.e,v.e,u.s) >= -EP)&&            (multiply(u.s,v.e,v.s)*multiply(v.e,u.e,v.s) >= -EP));}bool intersect_a(SEGMENT u,SEGMENT v){///两线段相交（不包括端点）    return ((intersect(u,v))&&            !online(u,v.s)&&            !online(u,v.e)&&            !online(v,u.e)&&            !online(v,u.s));}int lineintersect(LINE l1,LINE l2,POINT &p){///求两直线交点,有交点返回1和交点，没有返回0，重合返回2    double d = l1.a*l2.b-l2.a*l1.b;    double d2 = l1.a*l2.c-l2.a*l1.c;    double d3 = l1.b*l2.c-l2.b*l1.c;    if(fabs(d) < EP&&fabs(d2) < EP&&fabs(d3) < EP)return 2;    p.x = (l2.c*l1.b-l1.c*l2.b)/d;    p.y = (l2.a*l1.c-l1.a*l2.c)/d;    if(fabs(d) < EP)return 0;    return 1;}double dis(POINT a,POINT b){///求两点距离    return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));}///****************凸包**************************************int top;///凸包的顶点个数POINT p[maxn],stack[maxn];bool top_cmp(POINT a,POINT b){    double k = multiply(a,b,p[0]);    if(k > 0)return 1;    if(k < 0)return 0;    return dis(a,p[0]) < dis(b,p[0]);}void graham(int n){    top = 1;    int k = 0;    for(int i = 1;i < n;i ++)        if(p[i].y < p[k].y||((p[i].y == p[k].y)&&(p[i].x < p[k].x)))            k=i;    swap(p[0],p[k]);    sort(p+1,p+n,top_cmp);    stack[0] = p[0],stack[1] = p[1];    for(int i = 2;i < n;i ++){        while(top >= 1&&multiply(p[i],stack[top],stack[top-1]) >= 0)top --;        stack[++top] = p[i];    }}///******************凸包结束****************double get_area(int n,POINT v[]){///*********多边形面积    double ans;    if(n < 3)return 0;    ans = (double)(v[0].y*(v[n-1].x - v[1].x));    for(int i = 1;i < n;i ++){        ans += (double)(v[i].y*(v[i-1].x - v[i+1].x));    }    return ans/2.0;}double get_len(int n,POINT v[]){    double ans = dis(v[0],v[n-1]);    for(int i = 0;i < n-1;i ++){        ans += dis(v[i],v[i+1]);    }    return ans;}///*********************************************************************************POINT all[maxn];int main(){    //freopen("in.txt","r",stdin);    int n;    int cas = 1;    while(~scanf("%d",&n)){        if(n == 0)break;        for(int i = 0;i < n;i ++){            double x,y,v,l;            scanf("%lf%lf%lf%lf",&x,&y,&v,&l);            all[i].x = x;            all[i].y = y;            all[i].v = v;            all[i].l = l;        }        double min_val = 10000000;        double min_tree = 10000;        double num_wood = 0;        int status = 0;        for(int i = 1;i < (1<<n)-1;i ++){            int tot = 0;///the number of tree don't used            double val = 0;            double wood = 0;            for(int j = 0;j < n;j ++){                if(((1<<j)&i) == 0){                    p[tot++] = all[j];                }                else{                    val += all[j].v;//cout<<val<<endl;                    wood += all[j].l;                }            }//cout<<tot<<endl;            if(val > min_val)continue;            double len = 0;            if(tot == 1)len = 0;            else if(tot == 2)len = 2*dis(p[0],p[1]);            else {                graham(tot);                len = get_len(top+1,stack);            }            if(len <= wood){                if(val < min_val){                    min_val = val;                    min_tree = n - tot;                    num_wood = wood - len;                    status = i;                }                else if(val == min_val&&min_tree > n - tot){                    min_tree = n - tot;                    num_wood = wood - len;                    status = i;                }            }        }        printf("Forest %d\nCut these trees:",cas++);        bool used[maxn] = {0};        for(int i = 0;i < n;i ++){            if(((1<<i)&status))used[i] = 1;        }        for(int i = 0;i < n;i ++)            if(used[i]){                printf(" %d",i+1);            }        printf("\nExtra wood: %.2f\n\n",num_wood+EP);    }    return 0;}