Red and Black 深搜

B - Red and BlackCrawling in process...Crawling failedTime Limit:1000MSMemory Limit:30000KB 64bit IO Format:%I64d & %I64uSubmitStatusDescriptionThere 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 canmove only on black tiles.Write a program to count the number of black tiles which he can reach by repeating the moves described above.InputThe 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.OutputFor 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

题意：就是数一下一个@所在的位置可以走多少步的问题。

思路：找到@的地方进行深搜有四个方向，没有剪枝

AC代码：

#include<iostream>
#include<cstdio>
using namespace std;

char map[20][20];
int b[4]={0,1,-1,0};
int bb[4]={1,0,0,-1};
int r,c;
int sx,sy;
int tol;
int ok(int x,int y){
if(x<0||y<0||x>=r||y>=c||map[x][y]=='#')return 0;
return 1;
}
void dfs(int x,int y){
for(int i=0;i<4;i++){
map[x][y]='#';
if(ok(x+b[i],y+bb[i])){
tol++;
dfs(x+b[i],y+bb[i]);
}
}
}
int main (){
while(scanf("%d%d",&c,&r)!=EOF&&r!=0&&c!=0){
getchar();
for(int i=0;i<r;++i){
for(int j=0;j<c;j++){
scanf("%c",&map[i][j]);//记住此处gets（）是不行的，可能输入空格
if(map[i][j]=='@'){
sx=i;
sy=j;
}
}
if(i<r-1){getchar();}

}
tol=1;
dfs(sx,sy);
printf("%d\n",tol);
}
return 0;
}

这倒以前做过的，看见还是发现有点陌生，长时间不敲就忘了怎么弄，好好强化。。