ACM: poj 1964

City Game

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: 找出矩形形状的区域, 找出最大一个区域, 计算面积*3.

 

解题思路:

 1. 这类问题其实目的就一个, 算过的地方不想再重复计算, 降低时间复杂度. 那么就从第一行

       第一列开始. 一个一个往下计算, 同时我们记录每个点(i,j); 可以往上, 往左偏移和往右

       偏移到的位置(矩形扩张).

 2. 设up[i][j]: 表示当前(i,j)往上延伸最大长度; 如果(i,j区域空置)up[i][j] = up[i-1][j]+1;

    设leftSide[i][j]: 表示往左扩张; (i,j) = F; leftSide[i][j]=max(leftSide[i-1][j], lside+1);

    设rightSide[i][j]: 表示往右扩张; (i, j) = F; rightSide[i][j]=min(rightSide[i-1][j], rside-1)

 

代码:

#include
#include
#include
using namespace std;
#define MAX 1005

int n, m;
int g[MAX][MAX];
int up[MAX][MAX], leftSide[MAX][MAX], rightSide[MAX][MAX];

inline int min(int a, int b)
{
 return a < b ? a : b;
}

inline int max(int a, int b)
{
 return a > b ? a : b;
}

int main()
{
// freopen("input.txt", "r", stdin);
 int caseNum;
 scanf("%d", &caseNum);
 while(caseNum--)
 {
  scanf("%d %d", &n, &m);
  for(int i = 0; i < n; ++i)
  {
   for(int j = 0; j < m; ++j)
   {
    char ch[3];
    scanf("%s", ch);
    if(ch[0] == 'F') g[i][j] = 0;
    else g[i][j] = 1;
   }
  }
  
  int result = 0;
  for(int i = 0; i < n; ++i)
  {
   int lside = -1, rside = m;
   for(int j = 0; j < m; ++j)
   {
    if(g[i][j] == 1)
    {
     leftSide[i][j] = 0;
     up[i][j] = 0;
     lside = j;
    }
    else
    {
     up[i][j] = (i == 0 ? 1 : (up[i-1][j]+1));
     leftSide[i][j] = (i == 0 ? lside+1 : max(leftSide[i-1][j], lside+1));
    }
   }
   
   for(int j = m-1; j >= 0; --j)
   {
    if(g[i][j] == 1)
    {
     rightSide[i][j] = m;
     rside = j;
    }
    else
    {
     rightSide[i][j] = (i == 0 ? rside-1 : min(rightSide[i-1][j], rside-1));
     result = max(result, up[i][j]*(rightSide[i][j]-leftSide[i][j]+1));
    }
   }
  }
  printf("%d\n", result*3);
 }
 return 0;
}