题目代号：HDU 1312

题目链接：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): 20820    Accepted Submission(s): 12673

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

题目大意：一个男子站在‘.’上，他不能走到‘#’上问他能走到的‘.’的数量是多少。

解题思路：初始点bfs四个方向都遍历一次即可。

差点被自己气哭，第一次提交的时候忘记初始化数组，因为没有判断边界，直接导致WA。

AC代码：

# include <stdio.h>

# include <string.h>

# include <stdlib.h>

# include <iostream>

# include <fstream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <math.h>

# include <algorithm>

using namespace std;

# define pi acos(-1.0)

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define For(i,n,a) for(int i=n; i>=a; --i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define Fo(i,n,a) for(int i=n; i>a ;--i)

typedef long long LL;

typedef unsigned long long ULL;

const int MAXM=;

char a[MAXM][MAXM];

int n,m,ans;

int cx[]={-,,,};

int cy[]={,,-,};

struct node

{

    int x,y;

};

queue<node>Q;

void bfs()

{

    while(!Q.empty())

    {

        int x=Q.front().x;

        int y=Q.front().y;

        Q.pop();

        for(int i=;i<;i++)

        {

            int tx=x+cx[i];

            int ty=y+cy[i];

            if(a[tx][ty]=='.')

            {

                ans++;

                a[tx][ty]='#';

                Q.push(node{tx,ty});

            }

        }

    }

}

int main()

{

    //freopen("in.txt", "r", stdin);

    while(cin>>m>>n,n&&m)

    {

        mem(a,);

        for(int i=;i<=n;i++)

        {

            cin>>a[i]+;

            for(int j=;j<=m;j++)

            {

                if(a[i][j]=='@')

                {

                    Q.push(node{i,j});

                    a[i][j]='#';

                }

            }

        }

        ans=;

        bfs();

        cout<<ans<<endl;

    }

    return ;

}

HDU 1312 Red and Black（bfs，dfs均可，个人倾向bfs）的相关教程结束。