TOJ 2997: City Game

矫情吗；* 2022-05-10 14:50 155阅读 0赞

## **2997: City Game** ##

**描述**

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 �C 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.

**输入**

The first line of the input contains an integer K ?C 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 ?C reserved unit  
F ?C free unit  
In the end of each area description there is a separating line.

**输出**

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.

**样例输入**

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

**样例输出**

45  
0

这道题说是dp，但是我感觉是有点思维题的感觉？

可以先看看它的简单版↓ 有一个黄金矿工的题目是差不多的

[https://blog.csdn.net/qq\_38064038/article/details/78195045][https_blog.csdn.net_qq_38064038_article_details_78195045]

这道题求最大矩阵，可以建立一个up,left,right数组记录离他最近的R的位置，然后直接边做边算面积就好了~

#include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<iostream>
    #include<string>
    #include<algorithm>
    #include<map>
    #include<set>
    #include<queue>
    #include<vector>
    using namespace std;
    #define inf 0x3f3f3f3f
    #define LL long long
    char a[1005][1005];
    int u[1005][1005],l[1005][1005],r[1005][1005];
    int main()
    {
    	int t,n,m,i,j,ans;
    	char x;
    	scanf("%d",&t);
    	while(t--)
    	{
    		ans=-1;
    		scanf("%d%d",&n,&m);					
    		for(i=0;i<n;i++)
    		{
    			for(j=0;j<m;j++)
    			{
    				cin>>x;
    				a[i][j]=x;
    			}			
    		}		
    		for(i=0;i<n;i++)
    		{
    			int fl=-1,fr=m;
    			for(j=0;j<m;j++)
    			{
    				if(a[i][j]=='R')
    				{
    					u[i][j]=0;
    					l[i][j]=0;
    					fl=j;
    				}
    				else
    				{
    					if(i==0)
    					{
    						u[i][j]=1;
    						l[i][j]=fl+1;
    					}
    					else
    					{
    						u[i][j]=u[i-1][j]+1;
    						l[i][j]=max(l[i-1][j],fl+1);
    					}
    				}
    			}
    			for(j=m-1;j>=0;j--)
    			{
    				if(a[i][j]=='R')
    				{
    					r[i][j]=m;
    					fr=j;
    				}
    				else
    				{
    					if(i==0)
    					{
    						r[i][j]=fr-1;
    					}
    					else
    					{
    						r[i][j]=min(r[i-1][j],fr-1);
    					}
    					ans=max(ans,u[i][j]*(r[i][j]-l[i][j]+1));
    					//printf("%d\n",ans);
    				}
    			}
    			
    		}
    		/*for(i=0;i<n;i++)
            {
            	for(j=0;j<m;j++)
    	        	printf("%d ",u[i][j]);
            	printf("\n");
            }
    		printf("Up*****\n");
    		for(i=0;i<n;i++)
            {
            	for(j=0;j<m;j++)
    	        	printf("%d ",l[i][j]);
            	printf("\n");
            }
    		printf("Left*****\n");
    		for(i=0;i<n;i++)
            {
            	for(j=0;j<m;j++)
    	        	printf("%d ",r[i][j]);
            	printf("\n");
            }
    		printf("Right*****\n");*/
    		if(ans==-1)
    		ans=0;
    		printf("%d\n",ans*3);
    	}
    }

[https_blog.csdn.net_qq_38064038_article_details_78195045]: https://blog.csdn.net/qq_38064038/article/details/78195045