City Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2743    Accepted Submission(s): 1099


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
 
简单说一下题意：一块地，现在想找一个最大矩形。这个矩形必须全部由'F'构成。输出的则是最大矩形面积*3。如上述的例1。
 
话说这题我纠结了好久。知道要打表，不过思路很乱，不清楚该怎么打。问了一下别人后才明白过来。原来这个2维dp应该这么写才对。
 
 1 #include <stdio.h>
 2 #include <string.h>
 3 #define max 1010
 4 int dp[max][max], z[max], y[max];                      //dp是存放第一遍打表的值，z是存放左值，y是存放右值（zuo,you......）
 5 int Max(int a, int b)
 6 {
 7     return a > b ? a : b;
 8 }
 9 int main()
10 {
11     int T, l, r, i, j, square;
12     char s[10];
13     scanf("%d", &T);
14     while(T--)
15     {
16         square = -1;
17         memset(dp[0], 0, sizeof(dp[0]));              //这个赋零的方法是从别人那里学到的
18         scanf("%d%d", &l, &r);
19         for(i = 1; i <= l; i ++)
20         {
21             for(j = 1; j <= r; j ++)
22             {
23                 scanf("%s", &s[0]);
24                 if(s[0] == 'R')
25                     dp[i][j] = 0;
26                 else
27                     dp[i][j] = dp[i-1][j] + 1;       //测可行高度，也就是矩形的长度
28             }
29         }
30         for(i = 1; i <= l; i ++)
31         {
32             for(j = 1; j <= r; j ++)
33             {
34                 z[j] = j;
35                 y[j] = j;
36             }
37             dp[i][0] = dp[i][r+1] = -1;
38             for(j = 2; j <= r; j ++)
39             {
40                 while(dp[i][j] <= dp[i][z[j]-1])    //测左节点，也就是矩形的宽的一个端点
41                     z[j] = z[z[j]-1];
42             }
43             for(j = r-1; j >= 1; j --)
44             {
45                 while(dp[i][j] <= dp[i][y[j]+1])   //同理，另一个端点
46                     y[j] = y[y[j]+1];
47             }
48             for(j = 1; j <= r; j ++)
49             {
50                 dp[i][j] = dp[i][j] * (y[j] - z[j] + 1);
51                 if(square < dp[i][j])
52                     square = dp[i][j];            
53             }
54         }
55         printf("%d\n", square * 3);
56     }
57     return 0;
58 }

 

和这题类似的是1506，不过不需要加维度，比这个要简单不少http://acm.hdu.edu.cn/showproblem.php?pid=1506。