POJ 1873 The Fortified Forest

题意：是有n棵树，每棵的坐标，价值和长度已知，要砍掉若干根，用他们围住其他树，问损失价值最小的情况下又要长度足够围住其他树，砍掉哪些树。。             

思路：先求要砍掉的哪些树，在求剩下的树求凸包，在判是否可行。（枚举+凸包）               

 

// Time 407ms; Memory 200K

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#define inf 200000
using namespace std;

bool vis[15],tvis[15];
int n,m,minv,tn;

typedef struct point
{
	double x,y;
	int v;
	double l;
	int z;
	point(double xx=0,double yy=0,int vv=0,double ll=0,int zz=0):x(xx),y(yy),v(vv),l(ll),z(zz){}
}vector;

point p[15],ch[15];

bool operator < (point a,point b)
{
	return a.x<b.x || (a.x==b.x && a.y<b.y);
}
vector operator - (point a,point b)
{
	return vector(a.x-b.x,a.y-b.y);
}
double cross(vector a,vector b)
{
	return a.x*b.y-a.y*b.x;
}
double len(vector a)
{
	return sqrt(a.x*a.x+a.y*a.y);
}

int graph()
{
	int k,i;
	m=0;
	for(i=0;i<n;i++) if(!vis[i])
	{
		while(m>1 && cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
		ch[m++]=p[i];
	}
	k=m;
	for(i=n-2;i>=0;i--) if(!vis[i])
	{
		while(m>k && cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
		ch[m++]=p[i];
	}
	if(n>1) m--;
	double s1=0,s2=0;
	for(i=0;i<m;i++) s1+=len(ch[i]-ch[(i+1)%m]);
	for(i=0;i<n;i++) if(vis[i]) s2+=p[i].l;
	return s1<=s2;
}
void dfs(int d)
{
	int i,mv=0,mn=0;
	if(d==n)
	{
		if(graph())
		{
			for(i=0;i<n;i++) if(vis[i]) 
			{
				mv+=p[i].v;mn++;
			}
			if(mv<minv || (mv==minv && tn>mn)) 
			{
				for(i=0;i<n;i++) tvis[i]=vis[i];
				minv=mv;
				tn=mn;
			}
		}
		return;
	}
	dfs(d+1);
	vis[d]=1;
	dfs(d+1);
	vis[d]=0;
}
int main()
{
	int i,j,t=0,a[15];
	while(scanf("%d",&n)!=EOF && n)
	{
		for(i=0;i<n;i++)
		{
			scanf("%lf%lf%d%lf",&p[i].x,&p[i].y,&p[i].v,&p[i].l);
			p[i].z=i;
		}

		sort(p,p+n);
		minv=inf;tn=inf;
		memset(vis,0,sizeof(vis));

		dfs(0);
		if(t++) printf("
");
		printf("Forest %d
",t);
		printf("Cut these trees:");
		j=0;
		for(i=0;i<n;i++) if(tvis[i]) a[j++]=p[i].z+1;
		sort(a,a+j);
		for(i=0;i<j;i++) printf(" %d",a[i]);

		double s1=0,s2=0;
		for(i=0;i<n;i++) vis[i]=tvis[i];
		graph();
		for(i=0;i<m;i++) s1+=len(ch[i]-ch[(i+1)%m]);
		for(i=0;i<n;i++) if(vis[i]) s2+=p[i].l;
		printf("
Extra wood: %.2lf
",s2-s1);
	}
	return 0;
}