zoukankan      html  css  js  c++  java
  • HDOJ1312 Red and black(DFS深度优先搜索)

    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

    #include<stdio.h>
    #include<string.h>
    using namespace std;
    char s[50][50];
    int n,m;
    int dfs(int x,int y)
    {
        int i,j,k;
        if(x<0||x>n-1||y<0||y>m-1)
            return 0;
        
        if(s[x][y]=='#')
        {
            return 0;
        }    
        else
        {
            s[x][y]='#';
            return 1+dfs(x+1,y)+dfs(x-1,y)+dfs(x,y+1)+dfs(x,y-1);//搜索四个方向
        }
    }
    int main() 
    {
        int i,j,k;
        while(scanf("%d%d",&m,&n)!=EOF)
        {
            if(m==0&&n==0)
                break;
            for(i=0;i<n;i++)
            {
                scanf("%s",s[i]);
            }
            int sum=0;
            for(i=0;i<n;i++)
            {
                for(j=0;j<m;j++)
                {
                    if(s[i][j]=='@')
                    {
                        sum=dfs(i,j);    
                    } 
                } 
            }
            printf("%d
    ",sum);
        }
        return 0;
    }

    永不言弃!和ACM死磕到底!

    天晴了,起飞吧
  • 相关阅读:
    vue富文本编辑器
    vue图片上传组件
    vue全局使用axios插件请求ajax
    vue项目初始化时npm run dev报错webpack-dev-server解决方法
    vue axios使用form-data的形式提交数据
    react-keep-alive
    create-react-app 兼容 ie9
    next-定义路由
    next-支持css样式和按需加载antd
    react-错误边界
  • 原文地址:https://www.cnblogs.com/jianqiao123/p/10780499.html
Copyright © 2011-2022 走看看