POJ 1964 & 1494

 

City Game

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 3095		Accepted: 1181

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

Source

Southeastern Europe 2004

 

/* 1964 和1494基本是一样的 maxH[i]表示到当前行的第i列的最大子段 left[i]表示从left[i] + 1到第i - 1之间的所有列的当前最大子段都大于第i列的最大子段 right[i]表示从i + 1到right[i] - 1之间的所有列的当前最到子段都大于第i列的最大子段 即由第i列当前最大子段可以构成的举行的最大面积是(right[i] - left[i] - 1) * maxH[i] 时间复杂度是n ^ 2 */ #include <iostream> #define MAX_N 1005 #define MAXV(a, b) ((a) >= (b) ? (a) : (b)) #define MINV(a, b) ((a) <= (b) ? (a) : (b)) using namespace std; int maxH[MAX_N + 1]; int leftv[MAX_N + 1]; int rightv[MAX_N + 1]; int col, row, maxArea; int dpsolve() { int i, j; maxH[0] = maxH[col + 1] = INT_MIN; for(i = 1; i <= col; i++) { if(maxH[i] == 0) continue; for(j = i - 1; ; j = leftv[j]) { if(maxH[j] < maxH[i]) { leftv[i] = j; break; } } } for(i = col; i >= 1; i--) { if(maxH[i] == 0) continue; for(j = i + 1; ; j = rightv[j]) { if(maxH[j] < maxH[i]) { rightv[i] = j; break; } } } int curmax = INT_MIN; for(i = 1; i <= col; i++) { if(maxH[i] != maxH[i - 1]) { int curval = (rightv[i] - leftv[i] - 1) * maxH[i]; if(curval > curmax) curmax = curval; } } return curmax; } int main() { int caseN, i, j; char type; scanf("%d", &caseN); while(caseN--) { memset(maxH, 0, sizeof(maxH)); memset(leftv, 0, sizeof(leftv)); memset(rightv, 0, sizeof(rightv)); scanf("%d%d", &row, &col); maxArea = 0; for(i = 1; i <= row; i++) { for(j = 1; j <= col; j++) { cin>>type; if(type == 'F') maxH[j]++; else maxH[j] = 0; } int curval = dpsolve(); if(curval > maxArea) maxArea = curval; } printf("%d/n", maxArea * 3); } return 0; }