poj 1979 Red and Black_不见不散的结局是曲终人散_新浪博客

本文介绍了一道经典的迷宫遍历问题,通过深度优先搜索(DFS)算法来计算在一个由红色和黑色方块组成的矩形房间中,从指定起点出发能够到达的最大黑色方块数量。文章提供了完整的C++实现代码。

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Red and Black
Time Limit: 1000MSMemory Limit: 30000K
Total Submissions: 26216Accepted: 14238

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
题意:给你一个装‘#’,‘@’,‘.’的二维数组,从‘@’出发,‘.’代表可以走,‘#’代表不能走。求最多能走多少步;
思路:直接DFS吧,朝四个不同的方向走;
代码如下:
#include < cstdio >//为什么我的头文件发表出来就看不到了?
#include < cstring >
char s[25][25];
int m,n;
int i,j,ans;
int dfs(int i,int j)
{
    if(i<0||j<0||i>=n||j>=m)//可不能出界额;
        return 0;
    if(s[i][j]!='#')
    {
        s[i][j]='#';//走过的路,标记一下,下次不能再走;
        return 1+dfs(i+1,j)+dfs(i-1,j)+dfs(i,j+1)+dfs(i,j-1);//四个不同的方向;
    }
    else
        return 0;//保证走不通时就结束;
}
int main()
{
    while(~scanf("%d%d",&m,&n))
    {
        if(m==0&&n==0)
            break;
        for(i=0;i < n;i++)
            scanf("%s",&s[i]);
        for(i=0;i < n;i++)
            for(j=0;j < m;j++)
        {
            if(s[i][j]=='@')
                ans=dfs(i,j);
        }
        printf("%d\n",ans);
    }
    return 0;
}
 
 
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