HDU1312 BFS-Numerically Speaking

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).

题意：

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长方形地面铺有红砖和黑砖，人站在其中一块黑砖上，只能走黑砖不能走红砖，并且只能走前后左右的四块砖，不能往斜向走，问这个人能到达的黑砖有多少块；

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题解：

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简单的深度优先搜索问题，从起点开始向四个方向搜索，用一个队列保存是黑砖的点的坐标，接着往下面搜索，一直到队列为空，此时就可以遍历所有能够到达的店。

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#include <iostream>#include <queue>#include <string.h>#define MAXN 20+5using namespace std;char tile[MAXN][MAXN];bool vis[MAXN][MAXN];int w,h;int sum;//(dx,dy)表示四个方向const int dx[] = {0,1,0,-1};const int dy[] = {-1,0,1,0};queue<int> q;void bfs(int i, int j){    while(!q.empty()) q.pop();    q.push(i*h+j);//哈希法表示点的坐标    while(!q.empty())    {        int u = q.front();        q.pop();        //cout << u << endl;        int cx, cy;        cx = u/h;        cy = u%h;        //cout << cx << " " << cy << endl;        for(int k = 0; k < 4; k++)        {            int nx,ny;            nx = cx + dx[k];            ny = cy + dy[k];            if(nx >= 0 && nx < w && ny >= 0&& ny < h && tile[nx][ny] == '.' && vis[nx][ny] == false)            {                vis[nx][ny] = true;                q.push(nx*h+ny);                sum++;            }        }    }}int main(){    while(cin >> h >> w && h != 0 && w != 0)    {        int x0=0, y0=0;        memset(vis, 0, sizeof(vis));        sum = 1;        for(int i = 0; i < w; i++)            for(int j = 0; j < h; j++)            {                cin >> tile[i][j];                if(tile[i][j] == '@')                {                    x0 = i;                    y0 = j;                }            }        bfs(x0,y0);        //cout << x0 << " " << y0 << endl;        cout << sum << endl;    }    return 0;}