zoukankan      html  css  js  c++  java
  • A

    A - Red and Black(3.2.1)
    Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

    Description

    There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles. 

    Write a program to count the number of black tiles which he can reach by repeating the moves described above. 

    Input

    The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20. 

    There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows. 

    '.' - a black tile 
    '#' - a red tile 
    '@' - a man on a black tile(appears exactly once in a data set) 
    The end of the input is indicated by a line consisting of two zeros. 

    Output

    For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

    Sample Input

    6 9
    ....#.
    .....#
    ......
    ......
    ......
    ......
    ......
    #@...#
    .#..#.
    11 9
    .#.........
    .#.#######.
    .#.#.....#.
    .#.#.###.#.
    .#.#..@#.#.
    .#.#####.#.
    .#.......#.
    .#########.
    ...........
    11 6
    ..#..#..#..
    ..#..#..#..
    ..#..#..###
    ..#..#..#@.
    ..#..#..#..
    ..#..#..#..
    7 7
    ..#.#..
    ..#.#..
    ###.###
    ...@...
    ###.###
    ..#.#..
    ..#.#..
    0 0

    Sample Output

    45
    59
    6
    13
    #include<iostream>
    #include<cstring>
    using namespace std;
    int b,c,k,i,s,x,y,n;
    char a[21][21]; 
    int sousuo(int x,int y,int a1,int b)
    {
    	int n=0;
    	if(x>0)
    		if(a[x-1][y]=='.')
    		{n++;a[x-1][y]='a';n=n+sousuo(x-1,y,a1,b);}
        if(y>0)
    		if(a[x][y-1]=='.')
    		{n++;a[x][y-1]='a';n=n+sousuo(x,y-1,a1,b);}
    	if(x<a1)
          if(a[x+1][y]=='.')
    		{n++;a[x+1][y]='a';n=n+sousuo(x+1,y,a1,b);}
    	  if(y<b)
    		if(a[x][y+1]=='.')
    		{n++;a[x][y+1]='a';n=n+sousuo(x,y+1,a1,b);}
    		return n;
    }
    int main()
    {
    	int j;
    	while(cin>>k>>s&&k)
    	{
    		memset(a,'q',sizeof(a));
    	  for(i=0;i<s;i++)
    		  for(j=0;j<k;j++)
    		  {
    			  cin>>a[i][j];
    			  if(a[i][j]=='@')
    				  x=i,y=j;
    		  }
    
             n=sousuo(x,y,s,k);
    		 cout<<++n<<endl;
    	}
    	return 0;
    }


  • 相关阅读:
    BAT带队烧钱圈地华为们猛追云计算
    各浏览器的cookie的name个数/最大容量限制测试
    多备份:云端数据物流平台为企业提供云备份服务(通过增值服务盈利,数据备份相当于买保险)
    多备份CEO胡茂华:创业路上的五道坎
    蜡笔同步:同步通讯录,同步短信,用电脑发短信
    MIUI是小米的核心竞争力
    tggg
    Ubuntu 创建启动器
    TProcedure,TMethod,TNotifyEvent,TWndMethod的区别,并模拟点击按钮后发生的动作
    所有语言的Awesome
  • 原文地址:https://www.cnblogs.com/lxjshuju/p/7077689.html
Copyright © 2011-2022 走看看