hdu1255 覆盖的面积 线段树-扫描线

矩形面积并

线段树-扫描线裸题

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 #include<math.h>
 using namespace std;
 const int maxm=2e3+;
 const double eps=1e-;

 struct seg{
     double x,y1,y2;
     int c;
     bool operator < (const seg a)const{
         return x<a.x;
     }
 }s[maxm];

 double y[maxm];
 double st[maxm<<],st2[maxm<<];
 int cov[maxm<<];

 void pushup(int o,int l,int r){
     if(cov[o]>){
         st[o]=st2[o]=y[r]-y[l];
     }
     else if(cov[o]==){
         st[o]=y[r]-y[l];
         if(l+==r)st2[o]=;
         else st2[o]=st[o<<]+st[o<<|];
     }
     else{
         if(l+==r)st[o]=st2[o]=;
         else{
             st[o]=st[o<<]+st[o<<|];
             st2[o]=st2[o<<]+st2[o<<|];
         }
     }
 }

 void update(int o,int l,int r,seg a){
     if(y[l]>=a.y1&&y[r]<=a.y2){
         cov[o]+=a.c;
         pushup(o,l,r);
         return;
     }
     if(l+==r)return;
     int m=l+((r-l)>>);
     if(a.y1<=y[m])update(o<<,l,m,a);
     if(a.y2>y[m])update(o<<|,m,r,a);
     pushup(o,l,r);
 }

 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
         int n;
         scanf("%d",&n);
         memset(st,,sizeof(st));
         memset(cov,,sizeof(cov));
         memset(st2,,sizeof(st2));
         for(int i=;i<=n;++i){
             double x1,y1,x2,y2;
             scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
         /*    s[2*i-1].x=x1;
             s[2*i].x=x2;
             s[2*i-1].y1=y1;
             s[2*i].y1=y1;
             s[2*i-1].y2=y2;
             s[2*i].y2=y2;
             s[2*i-1].c=1;
             s[2*i].c=-1;
             y[2*i-1]=y1;
             y[2*i]=y2;*/
             s[*i-].x=x1;s[*i-].y1=y1;s[*i-].y2=y2;s[*i-].c=;
             y[*i-]=y1;
             s[*i].x=x2;s[*i].y1=y1;s[*i].y2=y2;s[*i].c=-;
             y[*i]=y2;
         }
         sort(y+,y+*n+);
         sort(s+,s+*n+);
 /*        int cnt=1;
         for(int i=2;i<=2*n;++i){
             if(y[i]!=y[i-1])y[++cnt]=y[i];
         }*/
         int cnt=*n;
         double ans=;
         update(,,cnt,s[]);
         for(int i=;i<=*n;++i){
             ans+=st2[]*(s[i].x-s[i-].x);
             update(,,cnt,s[i]);
         }
         printf("%.2lf\n",ans);
     }
     return ;
 }