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

.#.........

.#.#######.

.#.#.....#.

.#.#.###.#.

.#.#[email protected]#.#.

.#.#####.#.

.#.......#.

.#########.

...........

11 6

..#..#..#..

..#..#..#..

..#..#..###

..#..#..#@.

..#..#..#..

..#..#..#..

7 7

..#.#..

..#.#..

###.###

[email protected]

###.###

..#.#..

..#.#..

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=25;
char a[MAXM][MAXM];
int n,m,ans;
int cx[]={-1,1,0,0};
int cy[]={0,0,-1,1};

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=0;i<4;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,0);
        for(int i=1;i<=n;i++)
        {
            cin>>a[i]+1;
            for(int j=1;j<=m;j++)
            {
                if(a[i][j]==‘@‘)
                {
                    Q.push(node{i,j});
                    a[i][j]=‘#‘;
                }
            }
        }
        ans=1;
        bfs();
        cout<<ans<<endl;
    }
    return 0;
}