OpenJ_Bailian

题目：

OpenJ_Bailian - 1979 – Red and Black

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
45
59
6
13

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题意：

给出一个n*m的矩形，里面有红色砖块和黑色砖块，一个人站在黑色砖块上用@表示，他可以前后左右4种方向移动，求出他能到达的所有黑色砖块的个数。

分析：

用二维数组表示矩形，用dfs搜索，注意标记，如果红色砖块黑色砖块封住了的话，是不能进去的。

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代码如下：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;char ar[25][25];        //表示房间地板int n, m, cnt;int d[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};    //移动四个方向void dfs(int i, int j){    ar[i][j] = '!';        //访问过就变成！    //cout << i << " " << j << endl;    for(int k = 0; k < 4; k++)    {        int x = i + d[k][0];        int y = j + d[k][1];        if(x >= 0 && x < n && y >= 0 && y < m && ar[x][y] ！= '#')       //红色的不能过        {            if(ar[x][y] == '.')        //如果是没有访问过的就去访问            {                cnt++;                dfs(x, y);            }        }    }}int main(){    while(scanf("%d %d", &m, &n) != EOF && m + n)    {        int x, y;        getchar();        //注意吃掉字符！！！！        for(int i = 0; i < n; i++)        {            for(int j = 0; j < m; j++)            {                scanf("%c", &ar[i][j]);                if(ar[i][j] == '@')                {                    x = i;                    y = j;                }            }            getchar();        //吃掉字符！！！！！        }        cnt = 0;        dfs(x, y);        cout << cnt + 1 << endl;    }    return 0;}

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一定要注意吃掉字符……..本地都过不了都是因为这个…..最后好不容易才搞出来。