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  • Red and Black

    Red and Black

    Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
    Total Submission(s) : 49   Accepted Submission(s) : 30

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    Problem 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)

    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
    

    Source

    Asia 2004, Ehime (Japan), Japan Domestic
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int fang[4][2]={1,0,0,1,-1,0,0,-1};
    bool mapused[25][25];
    char map[25][25];
    int n,m;//列 行;
    int number;
    int dfs(int a,int b)
    {
        if(a>=0&&a<=m&&b>=0&&b<=n)
        {
            
              //判断有没有越界
               if(map[a][b]=='#')//遇墙 返回
               {
                    return 0;
               }
           else
               if(mapused[a][b]==1)//遇到走过的 返回
              {
                return 0;
               }
            else//没走过 且不是墙;
            {
                mapused[a][b]=1;
                number++;
                int k;
                for(k=0;k<4;k++)//搜索4个方向
                {
                    dfs(a+fang[k][0],b+fang[k][1]);
                }
        
        
            }
    
        }//越界结束
        else return 0;
    
    
    
    }
    int main()
    {
        
        
        
        while(scanf("%d %d",&n,&m)!=EOF)//列 行
        {
            if(m==0&&n==0) break;
            memset(mapused,0,sizeof(mapused));
            number=0;
            memset(map,'#',sizeof(map));//初始化
            int i,j;
            int starx,stary;
            for(i=0;i<m;i++)
            {
                
                for(j=0;j<n;j++)
                {
                    cin>>map[i][j];
                    if(map[i][j]=='@')
                    {
                        starx=i;//
                        stary=j;//
                        
                    
                    }
                
                }
            
            }//输入地图;
            dfs(starx,stary);
            printf("%d
    ",number);
            
    
        
        
        
        
        }
    
    
    
    
    
    }
    这题广搜 深搜都可以
    要注意初始化 和越界判断
    每走过一步就 用mapused[] 标记已走过的地方 避免重复走
    这题 起点重@开始 算能走多少格
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  • 原文地址:https://www.cnblogs.com/2013lzm/p/3263975.html
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