Uriel's Corner

Research Associate @ Harvard University / Research Interests: Computer Vision, Biomedical Image Analysis, Machine Learning
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POJ 1873 The Fortified Forest---凸包+枚举

Posted on 2010-05-05 21:04 Uriel 阅读(549) 评论(2)  编辑 收藏 引用 所属分类: POJ计算几何
        惭愧,这题从中午写到晚上,一开始一直卡着调也调不了,搞到晚上没办法换了个凸包模板,然后历经各种NC错误终于AC。。                                                          
        浙大模板凸包那里应该没什么问题,用过多少次了,应该是自己写挂了吧。。- -             
        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。。                                
         Discuss称这题为Final的水题。。于是我也水过去了。。枚举+构造凸包判可行。。代码如下。。凸包那里可以无视。。- -直接贴模板的。。
/*
Problem: 1873        User: Uriel
Memory: 192K        Time: 454MS
Language: C++        Result: Accepted
*/

#include
<math.h>
#include
<stdio.h>
#include
<stdlib.h>
#include
<algorithm>
using namespace std;

struct point
{
    
int flag;
    
double x,y,v,l;
}
;

point P[
100],convex[100],p1[100],p2[100],p[100];
double resl,suml,sumv,minv,ans;
int n,ansn;

double Dis(point a,point b)
{
    
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}


double multiply(const point& sp,const point& ep,const point& op) 
{
      
return((sp.x-op.x)*(ep.y-op.y)-(ep.x-op.x)*(sp.y-op.y));
}


bool cmp(point a,point b)
{
    
return a.flag < b.flag;
}


int partition(point a[],int p,int r) 
{
    
int i=p,j=r+1,k;
    
double ang,dis;
    point R,S;
    k
=(p+r)/2;
    R
=a[p];a[p]=a[k];a[k]=R;R=a[p];
    dis
=Dis(R,a[0]);
    
while(1
    
{
        
while(1
        
{
            
++i;
            
if(i>r) 
            
{
                i
=r;
                
break;
            }

            ang
=multiply(R,a[i],a[0]);
            
if(ang>0)
                
break;
            
else if(ang==0
            
{
                
if(Dis(a[i],a[0])>dis)
                
break;
            }

        }

        
while(1
        
{
             
--j;
            
if(j<p) 
            
{
                j
=p;
                
break;
            }

            ang
=multiply(R,a[j],a[0]);
            
if(ang<0)
                
break;
            
else if(ang==0
            
{
                
if(Dis(a[j],a[0])<dis)
                    
break;
            }

        }

        
if(i>=j)break;
        S
=a[i];
        a[i]
=a[j];
        a[j]
=S;
    }

    a[p]
=a[j];
    a[j]
=R;
    
return j;
}


void anglesort(point a[],int p,int r) 
{
    
if(p<r) 
    
{
        
int q=partition(a,p,r);
        anglesort(a,p,q
-1);
        anglesort(a,q
+1,r);
    }

}


/*PointSet传入散点,凸包上的点存在ch,n为点数,len为凸包顶点数*/
void Graham_scan(point PointSet[],point ch[],int n,int &len) 
{
    
int i,k=0,top=2;
    point tmp;
    
for(i=1;i<n;i++)
        
if ( PointSet[i].x<PointSet[k].x ||
            (PointSet[i].x
==PointSet[k].x) && (PointSet[i].y<PointSet[k].y))
               k
=i;
    tmp
=PointSet[0];
    PointSet[
0]=PointSet[k];
    PointSet[k]
=tmp;
    anglesort(PointSet,
1,n-1);
    
if(n<3
    
{
        len
=n;
        
for(int i=0;i<n;i++) ch[i]=PointSet[i];
        
return ;
    }

    ch[
0]=PointSet[0];
    ch[
1]=PointSet[1];
    ch[
2]=PointSet[2];
    
for (i=3;i<n;i++
    
{
        
while (multiply(PointSet[i],ch[top],ch[top-1])>=0) top--;
         ch[
++top]=PointSet[i];
    }

    len
=top+1;
}


int main()
{
    
int cse,g=1,i,j,k1,M,k2;
    
while(scanf("%d",&n),n)
    
{
        
for(i=0;i<n;i++)
        
{
            scanf(
"%lf %lf %lf %lf",&P[i].x,&P[i].y,&P[i].v,&P[i].l);
            P[i].flag
=i+1;
        }

        minv
=1e20;
        
for(i=0;i<(1<<n);i++)
        
{
            k1
=0;k2=0;
            sumv
=0;suml=0;
            
for(j=0;j<n;j++)
            
{
                
if(!(i & (1<<j)))
                
{
                    p1[k1].x
=P[j].x;
                    p1[k1].y
=P[j].y;
                    k1
++;
                }

                
else
                
{
                    p2[k2].flag
=P[j].flag;
                    sumv
+=P[j].v;
                    suml
+=P[j].l;
                    k2
++;
                }

            }

            Graham_scan(p1,convex,k1,M);
            resl
=0;
            
for(int j=0;j<M-1;j++)
            
{
                resl
+=Dis(convex[j],convex[j+1]);
            }

            resl
+=Dis(convex[0],convex[M-1]);
            
if(resl<=suml && sumv<minv)
            
{
                ans
=suml-resl;
                minv
=sumv;
                ansn
=k2;
                
for(j=0;j<k2;j++)p[j].flag=p2[j].flag;
            }

        }

        printf(
"Forest %d\n",g++);
        printf(
"Cut these trees:");
        
for(i=0;i<ansn;i++)if(p[i].flag!=p[i-1].flag)printf(" %d",p[i].flag);
        printf(
"\n");
        printf(
"Extra wood: %.2lf\n\n",ans);
    }

//    system("PAUSE");
    return 0;
}

Feedback

# re: POJ 1873 The Fortified Forest---凸包+枚举  回复  更多评论   

2012-03-20 20:24 by debuger
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# re: POJ 1873 The Fortified Forest---凸包+枚举  回复  更多评论   

2012-03-20 21:02 by Uriel
@debuger
谢谢提醒,原来没有考虑共线的情况,有共线时凸包那里会有bug
似乎改成这样就可以
while (multiply(PointSet[i], ch[top], ch[top - 1]) >= 0 && top >= 1) top--;
还有问题的话欢迎指教~

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