Red and Black（dfs，水题）

最新推荐文章于 2022-04-21 16:59:43 发布

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本文介绍了一种基于深度优先搜索的迷宫寻路算法。该算法在一个由黑色和红色方块组成的矩形房间中，从指定的起点出发，计算能够到达的最大黑色方块数量。通过递归方式遍历所有可达路径，并避免重复访问已检查过的节点。

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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

4559613

题解：从“@”开始走，”#“为墙，”.“可以走，问你最多走多少个地方。

代码如下：

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include<queue>
using namespace std;
const int MAX=1e2+10;
typedef long long LL;
char map[MAX][MAX];//记录地图
int n,m,t,vis[MAX][MAX],sx,sy,sum;
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};//上下左右移动
void dfs(int x,int y)
{

    //四种选择
    if(map[x+1][y]=='.'&&!vis[x+1][y])
    {
        sum++;//计数
        vis[x+1][y]=1;
        dfs(x+1,y);
    }
    if(map[x-1][y]=='.'&&!vis[x-1][y])
    {
        sum++;
        vis[x-1][y]=1;
        dfs(x-1,y);
    }
    if(map[x][y-1]=='.'&&!vis[x][y-1])
    {
        sum++;
        vis[x][y-1]=1;
        dfs(x,y-1);
    }
    if(map[x][y+1]=='.'&&!vis[x][y+1])
    {
        sum++;
        vis[x][y+1]=1;
        dfs(x,y+1);
    }
}
int main()
{
    while(cin>>m>>n)
    {
        if(n==0&&m==0)
            break;
        memset(vis,0,sizeof(vis));
        memset(map,'#',sizeof(map));//全部设为墙
        sum=0;
        for(int i=1; i<=n; i++)//从1开始，保证不越界
		{
			for(int j=1; j<=m; j++)
            {
                cin>>map[i][j];
            }
            getchar();
		}
		for(int i=1; i<=n; i++)
		{
			for(int j=1; j<=m; j++)
            {
            	if(map[i][j]=='@')
                {
                    sum++;
                    vis[i][j]=1;
                    dfs(i,j);
                    cout<<sum<<endl;
                    i=n;//结束循环
                    break;
                }
            }
		}
    }
    return 0;
}

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