using str use one space bsp 大於 money enc

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

題意：給出一個矩陣，上面是可以用的土地和不能用的土地，要在上面選一塊矩形的土地建房子，問最大能選多大的土地，土地每單位3塊大洋，最後輸出租金

核心是動態規劃做的。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int a[1010][1010];
 4 int l[1010],r[1010];
 5 int main()
 6 {
 7     int k;
 8     while(~scanf("%d",&k))
 9     {
10         int m,n;
11         while(k--)
12         {
13             scanf("%d %d",&m,&n);
14             memset(a,0,sizeof(a));
15             for(int i=0; i<m; i++)
16             {
17                 for(int j=0; j<n; j++)
18                 {
19                     char c[2];//以後輸入字符 中間帶空格的題目都可以直
20                     cin>>c;   // 接輸入字符數組 
21                     if(c[0]==‘F‘) a[i][j]=1;
22                 }
23             }
24             for(int i=1; i<m; i++)
25             {
26                 for(int j=0; j<n; j++)
27                 {
28                     if(a[i][j]!=0) a[i][j]=a[i-1][j]+1;
29                 }
30             }
31             int max=0;
32             for(int i=0; i<m; i++)//一行一行找過去，求最大面積 
33             {
34                 for(int j=0; j<n; j++)
35                 {
36                     l[j]=j;
37                     while(l[j]>0&&a[i][l[j]-1]>=a[i][j]) l[j]=l[l[j]-1];//向左邊,當前l[j]繼承符合要求的前 
38                 }                                                  // 一個的左邊界，動態規劃的核心，因為前一個 
39                 for(int j=n-1; j>-1; j--)              //點的高度大於當前點，所以前一個點的左邊界，當前點可以直接繼承使用 
40                 {
41                     r[j]=j;
42                     while(r[j]<n-1&&a[i][r[j]+1]>=a[i][j]) r[j]=r[r[j]+1];//向右邊，思路和左邊界一樣 
43                 }
44                 for(int j=0; j<n; j++)
45                 if(max<((r[j]-l[j]+1)*a[i][j])) max=((r[j]-l[j]+1)*a[i][j]);//表示從當前點向左右延伸的矩形面積 
46             }
47             printf("%d\n",max*3);//每單位面積3塊大洋
48 
49         }
50 
51     }
52     return 0;
53 }

hdu1505City Game（動態規劃）