POJ 1873 The Fortified Forest-优快云博客

本文提供了一个关于POJ1873“The Fortified Forest”问题的解决方案，使用了二进制枚举的方法来解决。代码中包含了完整的C++实现细节，包括凸包算法的运用以及如何通过枚举子集来找到最优解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

二进制枚举判断就行了。。。

一开始一直TLE不知道为什么，后来就莫名其妙地过了。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>

using namespace std;
#define FOR(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define DOR(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)

#define oo 1e6
#define eps 1e-8
#define nMax 100000
#define pb push_back

#define F first
#define S second

#define bug puts("OOOOh.....");
#define zero(x) (((x)>0?(x):-(x))<eps)

int dcmp(double x){
    if(fabs(x)<eps) return 0;
    return x>0?1:-1;
}
class point {
public:
    double x,y;
    point (double x=0,double y=0):x(x),y(y) {}
    void make(double _x,double _y) {x=_x;y=_y;}
    void read() { scanf("%lf%lf",&x,&y); }
    void out() { printf("%.2lf %.2lf\n",x,y);}
    double len() { return sqrt(x*x+y*y); }
    point friend operator - (point const& u,point const& v) {
        return point(u.x-v.x,u.y-v.y);
    }
    point friend operator + (point const& u,point const& v) {
        return point(u.x+v.x,u.y+v.y);
    }
    double friend operator * (point const& u,point const& v) {
        return u.x*v.y-u.y*v.x;
    }
    double friend operator ^ (point const& u,point const& v) {
        return u.x*v.x+u.y*v.y;
    }
    point friend operator * (point const& u,double const& k) {
        return point(u.x*k,u.y*k);
    }
	friend bool operator < (point const& u,point const& v){
		if(dcmp(v.x-u.x)==0) return dcmp(u.y-v.y)<0;
		return dcmp(u.x-v.x)<0;
	}
};
typedef class line{
public:
    point a,b;
    line() {}
    line (point a,point b):a(a),b(b){}
    void make(point u,point v) {a=u;b=v;}
    void read() { a.read(),b.read(); }
}segment;


inline int ConvexHull(point *p,int n,point *ret){
	sort(p,p+n);
	int m=0;
	for(int i=0;i<n;i++){
		while(m>1 && dcmp((ret[m-1]-ret[m-2])*(p[i]-ret[m-1]))<0) m--;
		ret[m++]=p[i];
	}
	int k=m;
	for(int i=n-2;i>=0;i--) {
		while(m>k && dcmp((ret[m-1]-ret[m-2])*(p[i]-ret[m-1]))<0) m--;
		ret[m++]=p[i];
	}
	if(n>1) m--;
	return m;
}
point q[nMax];
inline double check(point *p,int L,int n){
	int m=ConvexHull(p,n,q);
	double ret=0;
	for(int i=0;i<m;i++) ret += (q[i]-q[(i+1)%m]).len();
	return ret;
}

int n;
point P[nMax],p[nMax];
int l[nMax],v[nMax],cas=1;

void sovle(){
	int L=0,V=0,k=0;
	double ret,res;
	int ans,ansL,ansV;
	ans=0,ansL=0,ansV=0;
	for(int S=(1<<n)-1;S>=1;S--){
	    L=0;V=0;k=0;
	    int temp=S;
		for(int i=0;i<n;i++,temp>>=1) if(temp&1){
			L += l[i];
		}else {
			P[k++]=p[i];
			V += v[i];
		}
		if(V<ansV)continue;
		if(k>=2)
            ret = check(P,L,k);
        else
            ret=0;

		if(dcmp(ret-L)<=0){
            if(V>ansV){
                ans=S;ansV=V;ansL=n-k;res=L-ret;
            }else if(V==ansV){
                if(n-k<ansL){
                    ans=S;ansL=n-k;res=L-ret;
                }
            }
		}
	}
	printf("Forest %d\n",cas++);
	printf("Cut these trees:");
	for(int i=0;i<n;i++,ans>>=1) if(ans&1) printf(" %d",i+1);printf("\n");
	printf("Extra wood: %.2f\n",res);
}

int main(){
#ifndef  ONLINE_JUDGE
	freopen("input.txt","r",stdin);
#endif
    int first = 1;
	while(scanf("%d",&n),n){
	    if(first) first=0;
	    else printf("\n");
		FOR(i,0,n-1) {
			p[i].read();
			scanf("%d%d",&v[i],&l[i]);
		}
		sovle();
	}
	return 0;
}