HDU 1505 City Game

City Game

Time Limit: 2000/1000 MS (Java/Others)   

Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3015    Accepted Submission(s): 1236

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.  

View Code

 1 /*
 2   1506的升级版。一开始思路想错了，调了一个小时才调出来。
 3   首先遍历每一层，记录每一层中每个位置往上能达到的最大高度（这就构造出1506的模型了）
 4   lt[i][j],rt[i][j]记录的是第i层，第j个位置的左边界和右边界。
 5   循环查找该位置的最左边界和左右边界，则最大值为max(rt[i][j]-lt[i][j]+1)*height[i][j];
 6   
 7   109MS    13096K
 8   */
 9 #include <iostream>
10 #include <cstdio>
11 #include <cstring>
12 #include <algorithm>
13 #define SIZE 1005
14 
15 using namespace std;
16 
17 char map[SIZE][SIZE];
18 int lt[SIZE][SIZE],rt[SIZE][SIZE];
19 int height[SIZE][SIZE];
20 int T;
21 int n,m;
22 
23 int main()
24 {
25     scanf("%d",&T);
26     while(T--)
27     {
28         scanf("%d%d",&n,&m);
29         for(int i=1; i<=n; i++)
30         {
31             for(int j=1; j<=m; j++)
32             {
33                 cin >> map[i][j];
34                 lt[i][j] = rt[i][j] = j;
35             }
36         }
37         memset(height,0,sizeof(height));
38         for(int i=1; i<=m; i++)
39         {
40             if(map[1][i] == 'F')
41                 height[1][i] = 1;
42             else if(map[1][i] == '\0')
43                 break;
44         }
45         for(int i=2; i<=n; i++)
46         {
47             for(int j=1; j<=m; j++)
48             {
49                 if(map[i][j] == 'F')
50                     height[i][j] = height[i-1][j]+1;
51                 else if(map[i][j] == '\0')
52                     break;
53             }
54         }
55         for(int i=1; i<=n; i++)
56         {
57             height[i][0] = height[i][m+1] = -1;
58         }
59         for(int i=1; i<=n; i++)
60         {
61             for(int j=1; j<=m; j++)
62             {
63                 while(height[i][lt[i][j]-1] >= height[i][j])
64                     lt[i][j] = lt[i][lt[i][j]-1];
65             }
66         }
67         for(int i=1; i<=n; i++)
68         {
69             for(int j=m; j>=1; j--)
70             {
71                 while(height[i][rt[i][j]+1] >= height[i][j])
72                     rt[i][j] = rt[i][rt[i][j]+1];
73             }
74         }
75         int ans = 0;
76         for(int i=1; i<=n; i++)
77         {
78             for(int j=1; j<=m; j++)
79             {
80                 int temp = (rt[i][j]-lt[i][j]+1)*height[i][j];
81                 if(temp > ans)
82                     ans = temp;
83             }
84         }
85         printf("%d\n",ans*3);
86     }
87     return 0;
88 }