zoukankan      html  css  js  c++  java
  • 多校3 1004 Painter

    Painter

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 497    Accepted Submission(s): 239


    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.
     
    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
     
    Source
     
    Recommend
    wange2014   |   We have carefully selected several similar problems for you:  5325 5324 5323 5322 5321 
      1 #include <stdio.h>
      2 #include <string.h>
      3 #include <algorithm>
      4 using namespace std;
      5 int main()
      6 {
      7     int T;
      8     int n;
      9     int i,j,k;
     10     char a[55][56];
     11     scanf("%d",&T);
     12     while(T--)
     13     {
     14         scanf("%d",&n);
     15         for(i=0;i<n;i++)
     16         {
     17             scanf("%s",a[i]);
     18         }
     19         int l=strlen(a[0]);
     20 
     21         int ans=0,x,y,flg;
     22         for(i=0;i<l;i++)
     23         {
     24             flg=0;
     25              x=0,y=i;
     26             while(x<n && y<l)
     27             {
     28                 if(a[x][y]=='R' || a[x][y]=='G')
     29                 {
     30                     if(flg==0)
     31                         ans++,flg=1;
     32                 }
     33                 else
     34                 {
     35                     flg=0;
     36                 }
     37                 x++,y++;
     38             }
     39         }
     40         //printf("%d
    ",ans);
     41 
     42         for(i=1;i<l;i++)
     43         {
     44             flg=0;
     45             x=n-1,y=i;
     46             while(x>=0 && y<l)
     47             {
     48                 if(a[x][y]=='B' || a[x][y]=='G')
     49                 {
     50                     if(flg==0)
     51                         ans++,flg=1;
     52                 }
     53                 else
     54                 {
     55                     flg=0;
     56                 }
     57                 x--,y++;
     58             }
     59         }
     60         //printf("%d
    ",ans);
     61 
     62         for(j=1;j<n;j++)
     63         {
     64             flg=0;
     65             x=j,y=0;
     66             while(x<n && y<l)
     67             {
     68                 if(a[x][y]=='R' || a[x][y]=='G')
     69                 {
     70                     if(flg==0)
     71                         ans++,flg=1;
     72                 }
     73                 else
     74                 {
     75                     flg=0;
     76                 }
     77                 x++,y++;
     78             }
     79         }
     80         //printf("%d
    ",ans);
     81 
     82         for(j=0;j<n;j++)
     83         {
     84             flg=0;
     85             x=j,y=0;
     86             while(x>=0 && y<l)
     87             {
     88                 if(a[x][y]=='B' || a[x][y]=='G')
     89                 {
     90                     if(flg==0)
     91                         ans++,flg=1;
     92                 }
     93                 else
     94                 {
     95                     flg=0;
     96                 }
     97                 x--,y++;
     98             }
     99         }
    100         printf("%d
    ",ans);
    101     }
    102     return 0;
    103 }
    View Code
  • 相关阅读:
    C++数据类型与C#对应关系 c#调用WINDWOS API时,非常有用(转)
    Web应用系统中关闭Excel进程
    jquery下一个空格带来的血案
    导出Excel时发生COM组件失败的解决方案
    水晶报表的交叉表中增加超级链接
    JavaScript和ExtJS的继承 Ext.extend Ext.applyIf (转)
    SQL SERVER 2000数据库置疑处理
    PHP中对淘宝URL中ID提取
    树莓派+蓝牙适配器连接蓝牙设备
    树莓派摄像头模块转成H264编码通过RTMP实现Html输出
  • 原文地址:https://www.cnblogs.com/cyd308/p/4684911.html
Copyright © 2011-2022 走看看