zoukankan      html  css  js  c++  java
  • HDU 1505

    City Game

    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
     
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;
    int l[1005],r[1005],a[1005][1005];
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            int m,n;
            char s[15];
            scanf("%d%d",&m,&n);
            for(int i=1;i<=m;i++)
                for(int j=1;j<=n;j++)
                {
                    scanf("%s",s);
                    a[i][j]=(s[0]=='R'?0:1);
                }
            for(int i=2;i<=m;i++)
                for(int j=1;j<=n;j++)
                    if(a[i][j]==1)
                        a[i][j]+=a[i-1][j];
            int ans=0;
            for(int i=1;i<=m;i++)
            {
                for(int j=1;j<=n;j++)l[j]=j,r[j]=j;
                for(int j=2;j<=n;j++)
                {
                    if(a[i][j]<=a[i][j-1]&&a[i][j]!=0)
                    {
                        l[j]=l[j-1];
                        int k=j-1;
                        while(a[i][l[k]-1]>=a[i][j]&&k>1)k=l[k]-1;
                        l[j]=l[k];
                    }
                }
                for(int j=n-1;j>=1;j--)
                {
                    if(a[i][j]<=a[i][j+1]&&a[i][j]!=0)
                    {
                        r[j]=r[j+1];
                        int k=j+1;
                        while(a[i][r[k]+1]>=a[i][j]&&k<n)k=r[k]+1;
                        r[j]=r[k];
                    }
                }
                for(int j=1;j<=n;j++)
                    ans=max(ans,(r[j]-l[j]+1)*a[i][j]*3);
            }
            printf("%d
    ",ans);
        }
        return 0;
    }
  • 相关阅读:
    iphone界面详解
    Spring jdbcTemplate.queryForInt(sql)的奇怪问题,呵呵
    BCP 高效批量导入
    eclipse中javascript显示为乱码的解决办法
    spring jdbcTemplate返回RS
    Spring IOC DI 形象理解
    MOSS 2007 文档库事件处理
    showModalDialog和showModelessDialog使用心得
    XMLHTTP.open权限不够的解决
    体现JAVA中的面向对象思想,接口(抽象类)的用处 :饲养员给动物喂食物
  • 原文地址:https://www.cnblogs.com/homura/p/6709304.html
Copyright © 2011-2022 走看看