Red and Black(dfs模板)

本文介绍了一个基于深度优先搜索(DFS)算法的迷宫寻路问题解决方案。在一个由红色和黑色方块组成的矩形房间中,从一个黑色方块(起点)出发,仅能沿相邻的黑色方块移动,目标是计算能够到达的所有黑色方块的数量。通过递归地应用深度优先搜索,可以有效地遍历所有可达路径。

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Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 118 Accepted Submission(s): 92
 

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

 

Recommend

Eddy

#include <iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
char mp[21][21];
int dis[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};
int n,m;
int ans;
void dfs(int x,int y)
{
    ans++;
    mp[x][y]='#';
    for(int i=0; i<4; i++)
    {
        int tx=x+dis[i][0];
        int ty=y+dis[i][1];
        if(tx>=0&&tx<m&&ty>=0&&ty<n&&mp[tx][ty]=='.')
        {
            dfs(tx,ty);
        }
    }
    return;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)
            return 0;
        int x,y;
        for(int i=0; i<m; i++)
          scanf("%s",mp[i]);
        for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(mp[i][j]=='@')
                {
                    x=i;
                    y=j;
                }
            }
        }
        ans=0;
        dfs(x,y);
        printf("%d\n",ans);
    }
    return 0;
}

 

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