hdu1505City Game（动态规划）

City Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8359    Accepted Submission(s): 3630

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：给出一个矩阵，上面是可以用的土地和不能用的土地，要在上面选一块矩形的土地建房子，问最大能选多大的土地，土地每单位3块大洋，最后输出租金

核心是动态规划做的。

 #include<bits/stdc++.h>
 using namespace std;
 int a[][];
 int l[],r[];
 int main()
 {
     int k;
     while(~scanf("%d",&k))
     {
         int m,n;
         while(k--)
         {
             scanf("%d %d",&m,&n);
             memset(a,,sizeof(a));
             for(int i=; i<m; i++)
             {
                 for(int j=; j<n; j++)
                 {
                     char c[];//以后输入字符 中间带空格的题目都可以直
                     cin>>c;   // 接输入字符数组
                     if(c[]=='F') a[i][j]=;
                 }
             }
             for(int i=; i<m; i++)
             {
                 for(int j=; j<n; j++)
                 {
                     if(a[i][j]!=) a[i][j]=a[i-][j]+;
                 }
             }
             int max=;
             for(int i=; i<m; i++)//一行一行找过去，求最大面积
             {
                 for(int j=; j<n; j++)
                 {
                     l[j]=j;
                     while(l[j]>&&a[i][l[j]-]>=a[i][j]) l[j]=l[l[j]-];//向左边,当前l[j]继承符合要求的前
                 }                                                  // 一个的左边界，动态规划的核心，因为前一个
                 for(int j=n-; j>-; j--)              //点的高度大于当前点，所以前一个点的左边界，当前点可以直接继承使用
                 {
                     r[j]=j;
                     while(r[j]<n-&&a[i][r[j]+]>=a[i][j]) r[j]=r[r[j]+];//向右边，思路和左边界一样
                 }
                 for(int j=; j<n; j++)
                 if(max<((r[j]-l[j]+)*a[i][j])) max=((r[j]-l[j]+)*a[i][j]);//表示从当前点向左右延伸的矩形面积
             }
             printf("%d\n",max*);//每单位面积3块大洋

         }

     }
     return ;
 }