HDU - 1505 - City Game

先上题目：

City Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4101    Accepted Submission(s): 1710

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：找出最大的零矩阵结果等于最大零矩阵的面积*3。与HDU1506类似，只是将1506扩展成二维，需要求多次而已。

　　HDU - 1506 题解：http://www.cnblogs.com/sineatos/p/3600351.html

　　这里需要先求每一个位置的最大高度，然后再用1506的方法一行一行地求就可以了。

 

上代码：

 

#include <cstdio>
#include <cstring>
#define MAX 1002
#define max(x,y) (x > y ? x : y)
using namespace std;

bool map[MAX][MAX];
int cx[MAX][MAX];
int l[MAX],r[MAX];

int main()
{
    int n,m,t,maxn;
    char c[2];
    //freopen("data.txt","r",stdin);
    scanf("%d",&t);
    getchar();
    while(t--){
        scanf("%d %d",&n,&m);
        getchar();
        for(int i=1;i<=n;i++){
           for(int j=1;j<=m;j++){
                scanf("%s",c);
                if(c[0]=='R') map[i][j]=0;
                else map[i][j]=1;
           }
        }
        memset(cx,0,sizeof(cx));

        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(map[i][j]){
                    cx[i][j]=cx[i-1][j]+1;
                }
            }
        }

        maxn=0;
        for(int i=1;i<=n;i++){

            cx[i][0]=cx[i][m+1]=-1;

            for(int j=1;j<=m;j++){
                l[j]=r[j]=j;
            }

            for(int j=2;j<=m;j++){
                while(cx[i][j]<=cx[i][l[j]-1]){
                    l[j]=l[l[j]-1];
                }
            }

            for(int j=m-1;j>=1;j--){
                while(cx[i][j]<=cx[i][r[j]+1]){
                    r[j]=r[r[j]+1];
                }
            }

            for(int j=1;j<=m;j++){
                int ans=cx[i][j]*(r[j]-l[j]+1);
                maxn=max(ans,maxn);
            }
        }
        printf("%d\n",maxn*3);
    }
    return 0;
}