UVA 688-mobile Phone Coverage

Classic problem, rectangular area and.　     Solution: First, rectangular division, each rectangle two horizontal axis and two ordinate order, so that the 2n*2n interval, the interval to determine whether it is contained in the middle of n rectangles can be. Second, scan line. It hasn't been implemented yet. See: http://www.algorithmist.com/index.php/UVa_688 another: http://www.ii.uni.wroc.pl/boi/index.phtml?id=11

1#include <iostream>2#include <cstdio>3#include <algorithm>4 using namespacestd;5 6 intMain ()7 {8     intN, I, J, K, cases;9     DoubleArea, x[ the], y[ the], r[ the], dx[ the*2], dy[ the*2];TenCases =0; One      while(Cin>>n &&N) A     { -          for(i=0; i<n; i++) -         { theCIN >> X[i] >> Y[i] >>R[i]; -dx[i*2] = X[i]-r[i], dx[i*2+1] = X[i] +R[i]; -dy[i*2] = Y[i]-r[i], dy[i*2+1] = Y[i] +R[i]; -         } +Sort (DX, dx+2*n); -Sort (dy, dy+2*n); +Area =0.0; A          for(i =1; I <2*n; i++) at              for(j =1; J <2*n; J + +) -                  for(k =0; K < n; k++) -                 { -                     if(dx[i-1]>=x[k]-r[k] && dx[i]<=x[k]+R[k]) -                         if(dy[j-1]>=y[k]-r[k] && dy[j]<=y[k]+R[k]) -                         { inArea + = (Dx[i]-dx[i-1]) * (Dy[j]-dy[j-1]); -                              Break; to                         } +                 } -printf"%d%.2lf\n", ++cases, area); the     } *     return 0; $}

UVA 688-mobile Phone Coverage