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  • Code Forces 22B Bargaining Table

    B. Bargaining Table
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Bob wants to put a new bargaining table in his office. To do so he measured the office room thoroughly and drew its plan: Bob's office room is a rectangular room n × m meters. Each square meter of the room is either occupied by some furniture, or free. A bargaining table is rectangular, and should be placed so, that its sides are parallel to the office walls. Bob doesn't want to change or rearrange anything, that's why all the squares that will be occupied by the table should be initially free. Bob wants the new table to sit as many people as possible, thus its perimeter should be maximal. Help Bob find out the maximum possible perimeter of a bargaining table for his office.

    Input

    The first line contains 2 space-separated numbers n and m (1 ≤ n, m ≤ 25) — the office room dimensions. Then there follow n lines withm characters 0 or 1 each. 0 stands for a free square meter of the office room. 1 stands for an occupied square meter. It's guaranteed that at least one square meter in the room is free.

    Output

    Output one number — the maximum possible perimeter of a bargaining table for Bob's office room.

    Examples
    input
    3 3
    000
    010
    000
    
    output
    8
    
    input
    5 4
    1100
    0000
    0000
    0000
    0000
    
    output
    16
    
    
    暴力
    #include <iostream>
    #include <string.h>
    #include <stdlib.h>
    #include <algorithm>
    #include <math.h>
    #include <stdio.h>
    
    using namespace std;
    int n,m;
    char a[30][30];
    int main()
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%s",a[i]+1);
    
        bool tag=true;
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(a[i][j]=='1')
                    continue;
                //cout<<a[i][j]<<endl;
                for(int l=1;l+i-1<=n;l++)
                {
                    for(int r=1;r+j-1<=m;r++)
                    {
                        tag=true;
                        for(int k=i;k<=i+l-1;k++)
                        {
                            for(int p=j;p<=r+j-1;p++)
                            {
                                if(a[k][p]=='1')
                                   {
                                       tag=false;break;
                                   }
                            }
                            if(!tag)
                                break;
                        }
    
                        if(tag)
                           ans=max(ans,(l+r)*2);
                    }
    
                }
            }
        }
        printf("%d
    ",ans);
        return 0;
    
    }
    


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  • 原文地址:https://www.cnblogs.com/dacc123/p/8228717.html
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