hdoj1312Red and Black

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Red and Black

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

4559613
代码：
#include<stdio.h>#include<string.h>char map[1010][1010];int vis[1010][1010],n,m,ans,sx,sy;void dfs(int x,int y){if(x<0||x>=m||y<0||y>=n||vis[x][y]||map[x][y]=='#')return ;vis[x][y]=1;ans++;dfs(x+1,y);dfs(x,y+1);dfs(x-1,y);dfs(x,y-1);}int main(){while(scanf("%d%d",&n,&m)&&n||m){ans=0;memset(vis,0,sizeof(vis));memset(map,0,sizeof(map));for(int i=0;i<m;i++){scanf("%s",&map[i]);for(int j=0;j<n;j++){if(map[i][j]=='@')sx=i,sy=j;}}dfs(sx,sy);printf("%d\n",ans);}return 0;}
思路：深搜，dfs，这是第一次接触深搜题，乍一看一点思路也没有啊，然后学长讲完之后也是晕头转向的，迷迷糊糊，晕。然后自己照着代码敲了几遍，记住代码，再根据代码慢慢看懂了，memset(vis,0,sizeof(vis)); memset(map,0,sizeof(map));这个是要记住的，就是把所有的行行竖竖清零，在dfs，bfs里经常用到。然后就是定义dfs了，要记住确定方向dfs(x+1,y);dfs(x,y+1);dfs(x-1,y);dfs(x,y-1);，分别代表上下左右，其次就是找原始的点，然后本题就ac啦，注意不会多多敲几遍哈！！

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