BFS写HDU1312 Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25678    Accepted Submission(s): 15498

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

 

代码：

#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
int h,w,t;
int mp[21][21];
int dir[2][4]={ {1,-1,0,0},{0,0,1,-1} };
char s[21][21];
struct node
{
    int x,y;
};
queue<node> que;
void BFS(int x,int y)
{
    node u,v;
    mp[x][y]=1;
    u.x=x;
    u.y=y;
    que.push(u);
    while(!que.empty())
    {
        v=que.front();
        que.pop();
        for(int i=0;i<4;i++)
        {
            u.x=v.x+dir[0][i];
            u.y=v.y+dir[1][i];
            if(s[u.x][u.y]=='.' && !mp[u.x][u.y])
            {
                //BFS(u.x,u.y);
                mp[u.x][u.y]=1;
                que.push(u);
                t++;
            }
        }
    }

}
int main()
{
    int i,j;
    while(cin>>w>>h && h,w)
    {
        t=1;
        memset(s,0,sizeof(s));
        memset(mp,0,sizeof(mp));
        for(i=0;i<h;i++)
        for(j=0;j<w;j++)
        cin>>s[i][j];
        for(i=0;i<h;i++)
        {
            for(j=0;j<w;j++)
            if(s[i][j]=='@') break;
            if(s[i][j]=='@') break; 
        }
        BFS(i,j);
        cout<<t<<endl;
    }       
    return 0;
}