poj 1873 The Fortified Forest 搜索+凸包

链接

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题意：

给出一些树(2<=n<=15)，每个树有一个二维坐标，现在要砍下一些树，去做成围栏把其它树围起来。每棵树都有价值，还有砍了后能做成多少米的围栏。
为了节约材料，尽量使围栏的周长小。
现在要求使砍树的价值和最小，在此情况下要求砍树的数目最小。
要求输出砍掉哪些树，并且做完围栏后多余的材料可以围多少米。

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解法：

枚举所有状态（选中哪些树，不选哪些树），然后进行凸包。

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代码：

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])#define mem(a,x)  memset(a,x,sizeof a)#define ysk(x)  (1<<(x))#define sqr(x)  ((x)*(x))typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;const int maxn= 15    ;const double PI=acos(-1.0);const double eps=1e-10;int dcmp(double x){    if(fabs(x)<eps)  return 0;    else return x<0?-1:1;}struct Point{    double x,y;    int val,len,id;    Point(double x=0,double y=0):x(x),y(y) {};    bool operator ==(const Point B)const {return dcmp(x-B.x)==0&&dcmp(y-B.y)==0;}    bool operator<(const Point& b)const    {        return dcmp(x-b.x)<0|| dcmp(x-b.x)==0 &&dcmp(y-b.y)<0;    }};typedef Point Vector;Vector operator -(Vector A,Vector 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 Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}Vector operator +(Vector A,Vector B) {return Vector(A.x+B.x,A.y+B.y); }Vector operator *(Vector A,double p) {return Vector(A.x*p,A.y*p); }Vector operator /(Vector A,double p) {return Vector(A.x/p,A.y/p); }Vector operator -(Vector A)  {return  Vector(-A.x,-A.y);}double dis(Point A,Point B){    return sqrt(sqr(A.x-B.x)+sqr(A.y-B.y));}Point po[maxn+10],chp[maxn+10];bool use[maxn+4];int n;double ConvexHell(Point *p,int n,Point *ch,int rea){     int m=0;    for(int i=0;i<n;i++) if(use[i])    {        while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-1]   )<=0   )  m--;        ch[m++]=p[i];     }     int k=m;    for(int i=n-2;i>=0;i--) if(use[i])    {        while(m>k &&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-1] )<=0)   m--;        ch[m++]=p[i];    }    if(rea>1)  m--;//    if(m<=1)  return 0;//    if(m==2)  return dis(ch[0],ch[1]);    double D=0;    for0(i,m)    {        D+=dis(ch[i],ch[ (i+1)%m ]);    }//    cout<<"Convex "<<m<<endl;    return D;}void solve(){    int ed=ysk(n)-1;    int minVal=-1,num=-1,best=-1;    double extr=-1;    for0(s,ed+1)    {        int v=0,cnt=0,L=0;        for0(i,n)        {            if(s&ysk(i)) use[i]=1;            else  {use[i]=0;L+=po[i].len;v+=po[i].val,cnt++;}        }        double D=ConvexHell(po,n,chp,n-cnt);//        cout<<s<<" "<<D<<endl;        if(dcmp(L-D)<0)  continue;        if(minVal<0||minVal>v||minVal==v&&num>cnt )  {minVal=v;num=cnt;best=s;extr=L-D;}    }    printf("Cut these trees:");    vector<int >ve;    for0(i,n) if( !(ysk(i)&best) ) ve.push_back(po[i].id+1);    sort(all(ve));    for0(i,ve.size())  printf(" %d",ve[i]);    putchar('\n');    printf("Extra wood: %.2f\n\n",extr);}int main(){   std::ios::sync_with_stdio(false);   int kase=0;   while(cin>>n&&n)   {       for0(i,n) {cin>>po[i].x>>po[i].y>>po[i].val>>po[i].len;po[i].id=i;}       sort(po,po+n);      printf("Forest %d\n",++kase);       solve();   }   return 0;}