搜索--HDU 1312 Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22320    Accepted Submission(s): 13577


[align=left]Problem Description[/align]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.
 
[align=left]Input[/align]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)
 
[align=left]Output[/align]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).
 
[align=left]Sample Input[/align]

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0 [align=left]Sample Output[/align]

45
59
6
13

深搜代码

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int maxn=10010;
char a[22][22];
int f[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int l=1,s,aa,b,a1,b1;
int Judge(int x1,int y1)
{
if(x1<0||x1>=aa||y1<0||y1>=b)return 0;
if(a[x1][y1]!='.')return 0;
return 1;
}
void dfs(int x,int y)
{
for(int k=0;k<4;k++){
int x1,y1;
x1=x+f[k][0];
y1=y+f[k][1];
if(Judge(x1,y1)){
a[x1][y1]='#';
l++;
dfs(x1,y1);
l--;
s++;
}
}
}
int main(){
while(scanf("%d %d",&b,&aa)&&(b||aa)){
s=1;
for(int i=0;i<aa;i++){
for(int j=0;j<b;j++){
cin>>a[i][j];
if(a[i][j]=='@'){
a1=i;b1=j;
}
}
}
dfs(a1,b1);
cout<<s<<endl;
}
}