City Game - HDU 1505 最大内部矩形

解决城市游戏中的最大租金收益问题

最新推荐文章于 2018-11-25 17:11:04 发布

原创最新推荐文章于 2018-11-25 17:11:04 发布 · 659 阅读

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本博客探讨了在城市建造游戏中通过合理布局建筑物来最大化租金收益的问题。通过分析城市区域的网格布局，利用算法优化策略来确定最大面积的空地用于建设，从而实现租金的最大化。该过程涉及对现有建筑、树木、工厂和街道的考虑，以确保在不破坏现有设施的前提下进行建设。

City Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4468 Accepted Submission(s): 1884

Problem Description

Bob is a strategy game programming specialist. In his new city building game the gaming environment is as follows: a city is built up by areas, in which there are streets, trees,factories and buildings. There is still some space in the area that is unoccupied. The strategic task of his game is to win as much rent money from these free spaces. To win rent money you must erect buildings, that can only be rectangular, as long and wide as you can. Bob is trying to find a way to build the biggest possible building in each area. But he comes across some problems – he is not allowed to destroy already existing buildings, trees, factories and streets in the area he is building in.

Each area has its width and length. The area is divided into a grid of equal square units.The rent paid for each unit on which you're building stands is 3$.

Your task is to help Bob solve this problem. The whole city is divided into K areas. Each one of the areas is rectangular and has a different grid size with its own length M and width N.The existing occupied units are marked with the symbol R. The unoccupied units are marked with the symbol F.

Input

The first line of the input contains an integer K – determining the number of datasets. Next lines contain the area descriptions. One description is defined in the following way: The first line contains two integers-area length M<=1000 and width N<=1000, separated by a blank space. The next M lines contain N symbols that mark the reserved or free grid units,separated by a blank space. The symbols used are:

R – reserved unit

F – free unit

In the end of each area description there is a separating line.

Output

For each data set in the input print on a separate line, on the standard output, the integer that represents the profit obtained by erecting the largest building in the area encoded by the data set.

Sample Input

2
5 6
R F F F F F
F F F F F F
R R R F F F
F F F F F F
F F F F F F

5 5
R R R R R
R R R R R
R R R R R
R R R R R
R R R R R

Sample Output

45
0

题意：找到全为F的最大矩形面积，然后乘以3。

思路：记得以前挂过一道类似的，当是是用的n三次方复杂度，枚举两列，然后一维计算最大连续长度。但是这道题这样会T，正确的做法是对每个点先找到上方最大的长度，然后用优化的算法找出向左右扩展的最大长度。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int v[1010][1010],l[1010][1010],r[1010][1010],ans;
char s[10];
int main()
{ int t,n,m,i,j,k;
  scanf("%d",&t);
  while(t--)
  { scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
     for(j=1;j<=m;j++)
     { scanf("%s",s);
       l[i][j]=j;r[i][j]=j;
       if(s[0]=='F')
        v[i][j]=v[i-1][j]+1;
       else
        v[i][j]=0;
     }
     ans=0;
     for(i=1;i<=n;i++)
     { v[i][0]=-1;
       v[i][m+1]=-1;
       for(j=1;j<=m;j++)
        while(v[i][j]<=v[i][l[i][j]-1])
         l[i][j]=l[i][l[i][j]-1];
       for(j=m;j>=1;j--)
         while(v[i][j]<=v[i][r[i][j]+1])
          r[i][j]=r[i][r[i][j]+1];
       for(j=1;j<=m;j++)
         ans=max(ans,v[i][j]*(r[i][j]-l[i][j]+1));
     }
     printf("%d\n",ans*3);
  }
}