HDU 1312 Red and Black

关注

原创最新推荐文章于 2026-07-26 10:59:58 发布 2.5k 阅读 AI 写同款· GEO 优化 ›

DataWorks PyODPS第三方包全攻略：从打包到加载的完整流程本文详细介绍了在DataWorks的PyODPS节点中使用第三方Python包的完整流程，包括环境限制、打包工具pyodps-pack的使用、资源上传与加载实战，以及特殊场景处理技巧。通过本文，开发者可以掌握在MaxCompute受限环境中灵活使用自定义依赖的方法，提升数据处理效率。阅读详情

看了好多天的DFS，基本上的程序都是要借助解题报告才可以实现

但是这一题，确实依照自己的思路，完全靠自己写出来的，或许比较简单吧，所以特此发表下文章，犒劳一下自己，再接再厉

http://acm.hdu.edu.cn/showproblem.php?pid=1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2364 Accepted Submission(s): 1551

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
   
   
     
   
   
    #include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int direct[4][2] = { -1, 0, 1, 0, 0, 1, 0, -1 };  /*定义方向， 左右，上下*/

char map[21][21];               /*输入的字符串*/
bool mark[21][21];              /*标记走过的路程*/
bool flag;
int W, H;
int Dx, Dy;            //记录起始位置@，从这里开始进行搜索
int ans;                //记录满足的个数。初始化为1，因为@也包含在内
/*****底下是核心算法，主要是从
上下左右这四个方向进行搜索，注意
满足搜索的条件是不能越界，不能是#，
还有就是没有搜索过的--》主要是靠mark[i][j]
来实现*******/
void DFS( int x, int y )
{
    mark[x][y] = true;
    for( int k = 0; k < 4; k ++ )
    {
        int p = x + direct[k][0];
        int q = y + direct[k][1];
        if( p >= 0 && q >= 0 && p < H && q < W && !mark[p][q] && map[p][q] != '#' )
        {
            ans ++;
            DFS( p, q );
        }
    }
    return;
}

int main()
{
    int i, j, k;
    while( cin >> W >> H && ( W || H ) )   // W -> column, H -> row;
    {
        memset( mark, false, sizeof( mark ) );
        for( i = 0; i < H; i ++ )
            for( j = 0; j < W; j ++ )
            {
                cin >> map[i][j];
                if( map[i][j] == '@' )
                {
                    Dx = i;
                    Dy = j;
                }
            }
        ans = 1;
        DFS( Dx, Dy );
        cout << ans << endl;
    }
}