The E-pang Palace

Title: HDU5128 The E-pang Palace

Problem-Solving ideas: 1. Because the number of pillar is only 30, so the number of rectangular is up to 15*14/2=105, so the number of rectangular can be enumerated by brute force method.

2. Enumerate all the points as rectangular the lower left corner of each point.

3. Save the found rectangular to the lower-left corner and the upper-right corner x, y coordinates.

4. Make the found rectangular 22 to determine if the conditions are met, to find the largest area and.

5. Note that 2 rectangular nesting conditions are also eligible, and only the area of a large rectangular.

The code is as follows:

1#include <cstdio>2#include <cstring>3#include <algorithm>4 5 using namespacestd;6 7 structPillar8 {9     intx;Ten     inty; One}pillar[ *]; A  - structRect - { the     intX1; -     intY1; -     intx2; -     intY2; +}rect[ +]; -  + intmp[ About][ About]; A  at intMain () - { -     intN; -      while(SCANF ("%d", &n) &&N) -     { -         intL =0; inMemset (MP,0,sizeof(MP)); -         intI, J, K; to          for(i =0; I < n; i++) +         { -scanf"%d%d", &pillar[i].x, &pillar[i].y); theMP[PILLAR[I].X][PILLAR[I].Y] =1; *         } $          for(i =0; I < n; i++)Panax Notoginseng         { -              for(j = pillar[i].x +1; J <= $; J + +) the             { +                 if(Mp[j][pillar[i].y]) A                 { the                      for(k = pillar[i].y +1; K <= $; k++) +                     { -                         if(Mp[pillar[i].x][k] &&Mp[j][k]) $                         { $Rect[l].x1 =pillar[i].x; -Rect[l].y1 =pillar[i].y; -RECT[L].X2 =J; theRect[l].y2 =K; -l++;Wuyi                         } the                     } -                 } Wu             } -         } About         intsum =0; $          for(i =0; I < L; i++) -         { -              for(j = i +1; J < L; J + +) -             { A                 ints =0; +                 if((rect[i].x1-rect[j].x1) * (RECT[I].X2-RECT[J].X2) <0&& (rect[i].y1-rect[j].y1) * (Rect[i].y2-rect[j].y2) <0) the                 { -s = Max ((rect[i].y2-rect[i].y1) * (rect[i].x2-rect[i].x1), (rect[j].y2-rect[j].y1) * (RECT[J].X2-rect[j].x1)); $                 } the                 Else if(Rect[i].x1 > Rect[j].x2 | | rect[i].x2 < RECT[J].X1 | | rect[i].y1 > RECT[J].Y2 | | rect[i].y2 <rect[j].y1) the                 { thes = (rect[i].y2-rect[i].y1) * (rect[i].x2-rect[i].x1) + (rect[j].y2-rect[j].y1) * (RECT[J].X2-rect[j].x1); the                 } -sum =Max (sum, s); in             } the         } the         if(sum) About         { theprintf"%d\n", sum); the         } the         Else +         { -printf"imp\n"); the         }Bayi     } the  the  -     return 0; -}

AC is as follows:

The E-pang Palace