HDU1312——Red and Black （DFS)

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本文介绍了一种使用深度优先搜索（DFS）算法解决迷宫寻路问题的方法，通过遍历从起始位置可达的所有黑色方块来计算人可以到达的路径数量。提供了两种不同的实现方式，帮助读者理解如何在二维数组中应用DFS。

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12907 Accepted Submission(s): 7983

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

问题描述：'@'位置为人所在的位置，' . '为黑色块， ' # '为红色块，人儿只能走黑色块，问可以走多少块。

其实就是问@旁边有多少' . ' 是连着的

这个和HDU1241找油田那个题很像，不是很难，但是……呵！！！

AC代码

代码1：

和HDU1241做法一样

#include<iostream>
#include<cstdio>
#include<cstring>
#include <math.h>
#include <algorithm>
using namespace std;

char a[21][21];
int b[21][21],m,n,sum;
void dfs(int i,int j)
{
    if(a[i][j]=='#'||i<0||i>m-1||j<0||j>n-1||b[i][j]==1||a[i][j]!='.') 
        return ;
    b[i][j]=1;
    sum++;
    dfs(i+1,j);
    dfs(i-1,j);
    dfs(i,j-1);
    dfs(i,j+1);
}
int main()
{
    while(cin>>n>>m)
    {
        int x,y;
        if(m==0&&n==0) return 0;
        sum=0;
        for(int i=0; i<21; i++)
        {
            for(int j=0; j<21; j++)
            {
                b[i][j]=0;
            }
        }
        for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                cin>>a[i][j];
                if(a[i][j]=='@')
                {
                    x=i;
                    y=j;
                    a[i][j]='.';
                }
            }
        }
        dfs(x,y);
        /*for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                cout<<b[i][j];
            }
            cout<<endl;
        }*/
        cout<<sum<<endl;
    }
    return 0;
}

代码2：

#include<iostream>
#include<cstdio>
#include<cstring>
#include <math.h>
#include <algorithm>
using namespace std;

char a[25][25];
int m,n;
int sum;
void dfs(int i,int j)
{
    a[i][j]='!';///访问过的位置标记成！
    sum++;
    if(i-1>=0&&a[i-1][j]=='.')///各个方向
        dfs(i-1,j);
    if(i+1<m&&a[i+1][j]=='.')
        dfs(i+1,j);
    if(j-1>=0&&a[i][j-1]=='.')
        dfs(i,j-1);
    if(j+1>=0&&a[i][j+1]=='.')
        dfs(i,j+1);
}
int main()
{
    while(cin>>n>>m)
    {
        if(n==0&m==0)
            break;
        for(int i=0; i<m; i++) ///这个地方输入如果改成 for(int i=0; i<n; i++)
             cin>>a[i];        ///                          for(int j=0;j<m;j++)
                               ///                              cin>>a[i][j];
                               ///最后一组数据结果是错的，我也不知道为啥
        sum=0;
        for(int i=0; i<m; i++)
            for(int j=0; j<n; j++)
            {
                if(a[i][j]=='@') ///找人儿的位置，@
                    dfs(i,j);
            }
        cout<<sum<<endl;
    }
    return 0;
}