Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9902 Accepted Submission(s): 6158

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

Source

Asia 2004, Ehime (Japan), Japan Domestic

题意：

有一个长方形的房间布满了正方形的瓷砖，瓷砖要么红色要么黑色。一男子站在当中一块黑色瓷砖上，可向上下左右四个方向移动。但不能移动到红色瓷砖上，问他可到达的黑色瓷砖数量。

分析：

DFS搜索。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int dir[4][2] = {{-1, 0},{0, 1},{1, 0},{0, -1}};

int cnt, W, H;

char mp[21][21];

bool vis[21][21];

void dfs(int x, int y)

{

    vis[x][y] = true;

    for(int i = 0; i < 4; i++)

    {

        int tx = x + dir[i][0];

        int ty = y + dir[i][1];

        if(tx >= 1 && tx <= H && ty >= 1 && ty <= W && !vis[tx][ty] && mp[tx][ty] == '.')

        {

            cnt++;

            dfs(tx, ty);

        }

    }

}

int main()

{

    char c;

    int x, y;

    while(scanf("%d%d", &W, &H), W, H)

    {

        scanf("%c", &c);

        for(int i = 1; i <= H; i++)

        {

            for(int j = 1; j <= W; j++)

            {

                scanf("%c", &mp[i][j]);

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

                {

                    x = i;

                    y = j;

                }

            }

            scanf("%c", &c);

        }

        cnt = 1;

        memset(vis, false, sizeof(vis));

        dfs(x, y);

        printf("%d\n", cnt);

    }

    return 0;

}

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HDU1312 / POJ1979 / ZOJ2165 Red and Black（红与黑）解题报告