Ural1147 (Shaping Regions) rectangular cutting

 

A matrix of 10000*10000 with an initial color of 1 and a maximum of 2500 coloring times. The area of a rectangle is painted into a specific color, ask the final area of each color after painting.

 

Solution: inverted computing, rectangular cutting

 

Code:

/******************************************************* @author:xiefubao*******************************************************/#pragma comment(linker, /STACK:102400000,102400000)#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include 
     
      #include 
      
       #include #include 
       
        #include
        #include 
         
          #include 
          
           #include 
           
            //freopen (in.txt , r , stdin);using namespace std;#define eps 1e-8#define zero(_) (abs(_)<=eps)const double pi=acos(-1.0);typedef long long LL;const int Max=2510;const int INF=1e9+7;struct rec{ int x1,x2,y1,y2; int color;} recs[Max];int A,B;int n;int getans(int x1,int x2,int y1,int y2,int p){ while((p<=n)&&(x1>=recs[p].x2||x2<=recs[p].x1||y1>=recs[p].y2||y2<=recs[p].y1)) p++; if(p>n) return (x2-x1)*(y2-y1); int ans=0; if(x1
            
             recs[p].x2) ans+=getans(recs[p].x2,x2,y1,y2,p+1),x2=recs[p].x2; if(y2>recs[p].y2) ans+=getans(x1,x2,recs[p].y2,y2,p+1),y2=recs[p].y2; if(y1
             
              =1; i--) { int tool=getans(recs[i].x1,recs[i].x2,recs[i].y1,recs[i].y2,i+1); ans[recs[i].color]+=tool; ans[1]-=tool; } for(int i=1; i<=2500; i++) if(ans[i]) printf(%d %d,i,ans[i]); } return 0;}/*20 20 32 2 18 18 20 8 19 19 38 0 10 19 4*/