Red and Black （BFS）poj 1979

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 26181		Accepted: 14213

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)
The end of the input is indicated by a line consisting of two zeros.

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

题意为：求@可以到达.（点）的最多个数，别忘了@本身也算一个哦~！#代表不可以穿越！

大体就是这么个意思，不多说，直接上代码.......

#include <iostream>
#include <stdio.h>
#include <queue>
#include <string.h>
#define N 30
using namespace std;
int ans = 0, m, n;//ans用于计数的！
int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};//一点小技巧
int vis[N][N], map[N][N];//vis数组是用于标记是否访问过~
void bfs(int x, int y)
{
    int i,ty,tx,ttx,tty;
    queue<pair<int, int> > Q;//不会用queue的可以去百度看看  很好用滴
    Q.push(make_pair(x, y));
    vis[x][y] = 1;
    while(!Q.empty())
    {
        tx = Q.front().first, ty = Q.front().second;
        Q.pop();//别忘了一直要删除队头
        for(i = 0; i < 4; i++)
        {
            ttx = tx + dx[i], tty = ty + dy[i];
            if(ttx >= 0 && ttx < m && tty >= 0 && tty < n)
                if(!vis[ttx][tty] && !map[ttx][tty])
                {
                    vis[ttx][tty] = 1;
                    ans++;
                    Q.push(make_pair(ttx, tty));
                }
        }
    }
}
int main()
{
    int i, j, x, y;
    while(~scanf("%d%d", &n, &m))
    {
        ans = 1;
        memset(vis, 0, sizeof(vis));//比for循环清零好使
        getchar();//有些人输入老是出问题，看看是不是这的原因
        if(n == 0 && m == 0)
            break;
        char s[N];
        for(i = 0; i < m; i++)
        {
            gets(s);
            for(j = 0; j < n; j++)
            {
                if(s[j] == '.')
                    map[i][j] = 0;
                else if(s[j] == '#')
                    map[i][j] = 1;
                else
                    x = i, y = j;//@的坐标
            }
        }
        bfs(x, y);
        printf("%d
", ans);
    }
    return 0;
}