zoukankan      html  css  js  c++  java
  • HDU 5319

    Painter

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 816    Accepted Submission(s): 376


    Problem Description
    Mr. Hdu is an painter, as we all know, painters need ideas to innovate , one day, he got stuck in rut and the ideas dry up, he took out a drawing board and began to draw casually. Imagine the board is a rectangle, consists of several square grids. He drew diagonally, so there are two kinds of draws, one is like ‘’ , the other is like ‘/’. In each draw he choose arbitrary number of grids to draw. He always drew the first kind in red color, and drew the other kind in blue color, when a grid is drew by both red and blue, it becomes green. A grid will never be drew by the same color more than one time. Now give you the ultimate state of the board, can you calculate the minimum time of draws to reach this state.
     

    Input
    The first line is an integer T describe the number of test cases.
    Each test case begins with an integer number n describe the number of rows of the drawing board.
    Then n lines of string consist of ‘R’ ‘B’ ‘G’ and ‘.’ of the same length. ‘.’ means the grid has not been drawn.
    1<=n<=50

    The number of column of the rectangle is also less than 50.


    Output
    Output an integer as described in the problem description.

     
    Sample Input
    2
    4
    RR.B
    .RG.
    .BRR
    B..R
    4
    RRBB
    RGGB
    BGGR
    BBRR
     


    Sample Output
    3

    6

    //遍历一遍地图 当碰到R的时候这一斜对角线都变为'.' 当碰到G的时候变为B

    // 由于R和B 的方向不同 要推断两次  对角线有可能没有所有画完  一条对角线有可能画多次 

    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    using namespace std;
    char s[60][60];
    
    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)
        {
            int n;
            scanf("%d",&n);
            getchar();
            for(int i=0;i<n;i++)
                gets(s[i]);
            int ans=0;
            int len=strlen(s[0]);
            for(int i=0;i<n;i++)
            {
                for(int k=0;k<len;k++)  //注意长度 
                {
                    int t=k;
                    if((s[i][k]=='R')||(s[i][k]=='G'))
                    {
                        ans++;
                        for(int j=i;j<n&&t<len;j++)
                        {
    
                            if(s[j][t]=='.'||s[j][t]=='B')
                                break;
                            if(s[j][t]=='G')
                                s[j][t]='B';
                            else if(s[j][t]=='R')
                                s[j][t]='.';
                            t++;
                        }
                    }
                    t=k;
                    if((s[i][k]=='B')||(s[i][k]=='G'))
                    {
                        ans++;
                        for(int j=i;j<n&&t>=0;j++)
                        {
    
                            if(s[j][t]=='.'||s[j][t]=='R')
                                break;
                            if(s[j][t]=='G')
                                s[j][t]='R';
                            else if(s[j][t]=='B')
                                s[j][t]='.';
                            t--;
                        }
                    }
                }
    
            }
            printf("%d
    ",ans);
        }
        return 0;
    }


  • 相关阅读:
    centos 新增用户, 然后他在主目录添加网站403Forbbiden
    linux 把用户加入一个组&从这个组中移除
    [THINKPHP] 温故知新之getFieldBy
    php 获取指定月份的开始结束时间
    apache 占用内存总量与每个apache进程的平均内存占用量计算
    网站并发300就很慢
    centos定时备份数据库超简单示例
    php导出excel时间错误(同一个时间戳,用date得到不同的时间)
    设置iframe 载入页面的效果跟直接打开这个页面一样
    node基础09:第2个node web服务器
  • 原文地址:https://www.cnblogs.com/brucemengbm/p/7066064.html
Copyright © 2011-2022 走看看