HDU1505 最大全零矩阵

城市建筑策略：最大全零矩形

最新推荐文章于 2022-07-23 21:26:00 发布

原创最新推荐文章于 2022-07-23 21:26:00 发布 · 3.5k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#integer #each #input #output #behavior #build

动态规划DP 专栏收录该内容

45 篇文章

订阅专栏

在Bob的城市建设游戏中，玩家需在有限的空间内建造最大的矩形建筑以获得租金收入。每个区域是一个网格，其中'F'代表空地，'R'代表已占用。目标是找出每个区域中能容纳的最大矩形空地，以此计算最大收益。题目提供了输入输出样例及程序代码，涉及二维矩阵问题和动态规划解决方法。

City Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1241 Accepted Submission(s): 448

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

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

5 5

R R R R R

Sample Output

HDU1056 的加强版 , 在一个 01 矩阵里面找一个最大的全 0 矩阵 , 先求出每个 0 能够向上找到的 0 的个数 , 然后以 HDU1056 的形式进行 DP.

#include<stdio.h> #include<string.h> #define N 1005 int m,n,T,i,j,u[N][N],l[N][N],r[N][N],t,maxs; char g[N][N],c[10]; int main() { scanf("%d",&T); while (T--) { scanf("%d%d",&m,&n); for (i=1;i<=n;++i) u[0][i]=0; for (i=1;i<=m;++i) { for (j=1;j<=n;++j) { scanf("%s",c); if (c[0]=='F') u[i][j]=u[i-1][j]+1; else u[i][j]=0; } } maxs=-1; for (i=1;i<=m;++i) { u[i][0]=u[i][n+1]=-1; for (j=1;j<=n;++j) l[i][j]=r[i][j]=j; for (j=2;j<=n;++j) { while (u[i][l[i][j]-1]>=u[i][j]) l[i][j]=l[i][l[i][j]-1]; } for (j=n-1;j>=1;--j) { while (u[i][r[i][j]+1]>=u[i][j]) r[i][j]=r[i][r[i][j]+1]; } for (j=1;j<=n;++j) { if (u[i][j]*(r[i][j]-l[i][j]+1)>maxs) maxs=u[i][j]*(r[i][j]-l[i][j]+1); } } printf("%d/n",maxs*3); } return 0; }