• POJ1264 SCUD Busters 凸包


    POJ1264

    有m个国家(m<=20)对每个国家给定n个城镇 这个国家的围墙是保证围住n个城镇的周长最短的多边形 必然是凸包

    进行若干次导弹发射 落到一个国家内则国家被破坏

    最后回答总共有多少面积被破坏

    首先求凸包

    然后判断点是否在凸包内 要用O(logn)的判断方法 不然会超时

    这道题常数卡的有点紧 TLE三次才过

    蒟蒻没救了 2017年做1991年的题还被卡常数

    #include<iostream>
    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    #include<math.h>
    #include<algorithm>
    #include<queue>
    #include<vector>
    using namespace std;
    
    const double eps=1e-9;
    
    int cmp(double x)
    {
     if(fabs(x)<eps)return 0;
     if(x>0)return 1;
     	else return -1;
    }
    
    const double pi=acos(-1.0);
    
    inline double sqr(double x)
    {
     return x*x;
    }
    
    
    
    
    
    
    struct point
    {
     double x,y;
     point (){}
     point (double a,double b):x(a),y(b){}
     void input()
     	{
     	 scanf("%lf%lf",&x,&y);
    	}
     friend point operator +(const point &a,const point &b)
     	{
     	 return point(a.x+b.x,a.y+b.y);
    	}	
     friend point operator -(const point &a,const point &b)
     	{
     	 return point(a.x-b.x,a.y-b.y);
    	}
     friend bool operator ==(const point &a,const point &b)
     	{
     	 return cmp(a.x-b.x)==0&&cmp(a.y-b.y)==0;
    	}
     friend point operator *(const point &a,const double &b)
     	{
     	 return point(a.x*b,a.y*b);
    	}
     friend point operator*(const double &a,const point &b)
     	{
     	 return point(a*b.x,a*b.y);
    	}
     friend point operator /(const point &a,const double &b)
     	{
     	 return point(a.x/b,a.y/b);
    	}
     double norm()
     	{
     	 return sqrt(sqr(x)+sqr(y));
    	}
    };
    
    struct line
    {
     point a,b;
     line(){};
     line(point x,point y):a(x),b(y)
     {
     	
     }
    };
    double det(const point &a,const point &b)
    {
     return a.x*b.y-a.y*b.x;
    }
    
    double dot(const point &a,const point &b)
    {
     return a.x*b.x+a.y*b.y; 
    }
    
    double dist(const point &a,const point &b)
    {
     return (a-b).norm();
    }
    
    point rotate_point(const point &p,double A)
    {
     double tx=p.x,ty=p.y;
     return point(tx*cos(A)-ty*sin(A),tx*sin(A)+ty*cos(A));
    }
    
    
    
    
    bool parallel(line a,line b)
    {
     return !cmp(det(a.a-a.b,b.a-b.b));
    }
    
    bool line_joined(line a,line b,point &res)
    {
     if(parallel(a,b))return false;
     double s1=det(a.a-b.a,b.b-b.a);
     double s2=det(a.b-b.a,b.b-b.a);
     res=(s1*a.b-s2*a.a)/(s1-s2);
     return true;
    }
    
    bool pointonSegment(point p,point s,point t)
    {
     return cmp(det(p-s,t-s))==0&&cmp(dot(p-s,p-t))<=0;
    }
    
    void PointProjLine(const point p,const point s,const point t,point &cp)
    {
     double r=dot((t-s),(p-s))/dot(t-s,t-s);
     cp=s+r*(t-s);
    }
    
    
    struct polygon_convex
    {
     vector<point>P;
     polygon_convex(int Size=0)
     	{
     	 P.resize(Size);
    	}	
    };
    
    bool comp_less(const point &a,const point &b)
    {
     return cmp(a.x-b.x)<0||cmp(a.x-b.x)==0&&cmp(a.y-b.y)<0;
     
    }
    
    
    polygon_convex convex_hull(vector<point> a)
    {
     polygon_convex res(2*a.size()+5);
     sort(a.begin(),a.end(),comp_less);
     a.erase(unique(a.begin(),a.end()),a.end());//删去重复点 
     int m=0;
     for(int i=0;i<a.size();i++)
     	{
     	 while(m>1&&cmp(det(res.P[m-1]-res.P[m-2],a[i]-res.P[m-2]))<=0)--m;
     	 res.P[m++]=a[i];
    	}
     int k=m;
     for(int i=int(a.size())-2;i>=0;--i)
     	{
     	 while(m>k&&cmp(det(res.P[m-1]-res.P[m-2],a[i]-res.P[m-2]))<=0)--m;
     	 res.P[m++]=a[i];
    	}
     res.P.resize(m);
     if(a.size()>1)res.P.resize(m-1);
     return res;
    }
    
    bool is_convex(vector<point> &a)
    {
     for(int i=0;i<a.size();i++)
     	{
     	 int i1=(i+1)%int(a.size());
     	 int i2=(i+2)%int(a.size());
     	 int i3=(i+3)%int(a.size());
     	 if((cmp(det(a[i1]-a[i],a[i2]-a[i1]))*cmp(det(a[i2]-a[i1],a[i3]-a[i2])))<0)
    	  	return false;
    	}
     return true;
    }
    int containO(const polygon_convex &a,const point &b)
    {
     int n=a.P.size();
     point g=(a.P[0]+a.P[n/3]+a.P[2*n/3])/3.0;
     int l=0,r=n;
     while(l+1<r)
     	{
     	 int mid=(l+r)/2;
     	 if(cmp(det(a.P[l]-g,a.P[mid]-g))>0)
     	 	{
     	 	 if(cmp(det(a.P[l]-g,b-g))>=0&&cmp(det(a.P[mid]-g,b-g))<0)r=mid;
     	 	 	else l=mid;
    		}else
    			{
    			 if(cmp(det(a.P[l]-g,b-g))<0&&cmp(det(a.P[mid]-g,b-g))>=0)l=mid;
     	 	 		else r=mid;	
    			}
    	} 
     r%=n;
     int z=cmp(det(a.P[r]-b,a.P[l]-b))-1;
     if(z==-2)return 1;
     return z;	
    }
    
    polygon_convex pc[30];
    double area(int n)
    {
     double ans=0;
     vector<point>a;
     a.clear();
     for(int i=0;i<pc[n].P.size();i++)
        a.push_back(pc[n].P[i]);
     a.push_back(a[0]);
     for(int i=0;i<a.size()-1;i++)ans+=det(a[i+1],a[i]);
     //cout<<ans/2<<endl;
     return ans/2.0;
    }
    
    int tot;
    double are[30];
    
    vector<point> pp;
    int shoot[600][600];
    bool damed[30];
    int main()
    {//freopen("t.txt","r",stdin);
     int n;
     tot=1;
     while(scanf("%d",&n)&&n!=-1)
     	{
    	 pp.clear();
     	 for(int i=0;i<n;i++)
     	 	{
     	 	 point p;
     	 	 p.input();
     	 	 pp.push_back(p);
    		}
    	 pc[tot]=convex_hull(pp); 
    	 //cout<<pc[tot].P.size()<<endl;
    	 are[tot]=area(tot);
    	 //cout<<are[tot]<<endl;
    	 tot++;
    	}		
     int x,y;
     memset(damed,0,sizeof(damed));
     int sum=0;
     memset(shoot,0,sizeof(shoot));
     while(scanf("%d%d",&x,&y)!=EOF)
     	{
     	 if(sum==tot-1)break;
     	 if(shoot[x][y]==0)
     	 	{
     	 	 for(int i=1;i<tot;i++)
    		   	if(containO(pc[i],point(x,y)))
    			   	{
    			   	 shoot[x][y]=i;
    			   	 break;
    				}	
    		}
     	 if(!damed[shoot[x][y]]){sum++;damed[shoot[x][y]]=1;}
    	}
     double tarea=0;
     for(int i=1;i<tot;i++)
     	{
     	 //cout<<are[i]<<endl;
     	 if(damed[i])tarea+=are[i];
    	}
     printf("%.2lf
    ",-tarea+0.0005);
     return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/heisenberg-/p/6691582.html
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