pojRed and Black(dfs)

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 27213 Accepted: 14782

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

4559613

题目大意：‘#’表示红方块，‘.’表示黑方块，某人初始位置在‘@',可以向上下左右四个方向走，他只能走黑块；问从起点他能走的黑块的个数共有多少？

解题思路：本体可以用深搜来解决；算是比较基础；主要思路就是从起点开始上下左右开始搜索，并不断将走过的黑块标记为红块，表示已走过；此过程定义一个全局变量来统计走过的黑块数；当所有可以走的黑块都被走一遍，搜索完毕。

AC代码：

//poj-1979dfs #include<stdio.h>#include<string.h>char map[30][30];int n,m,sx,sy,cnt;void dfs(int x,int y){if(x<0||x>=m||y<0||y>=n) return ;  //判断是否越界； if(map[x][y]=='#') return ;        //红块不能走； map[x][y]='#';cnt++;          //统计走过的黑块数 dfs(x-1,y);     //四个方向开始搜索 dfs(x+1,y);dfs(x,y+1);dfs(x,y-1);}int main(){   int i,j,k;   while(scanf("%d%d",&n,&m)&&(n|m))   {    cnt=0;    memset(map,0,sizeof(map));     for(i=0;i<m;i++)     {     getchar();     scanf("%s",map[i]);     for(j=0;j<n;j++)     {     if(map[i][j]=='@'){     sx=i;sy=j; } } } dfs(sx,sy); printf("%d\n",cnt);   }   return 0;}