HDOJ1312 Red and Black 深搜dfs

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 15340 Accepted Submission(s): 9498



[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

#include<iostream>
#include<cstdio>
using namespace std;
char str[25][25];
int w,h;
int dfs(int a,int b)
{
if(str[a][b]=='#'||a>=h||a<0||b<0||b>=w)//到边界就结束，闪人
return 0;
else
{
str[a][b]='#';//走过的地方都更新为墙--‘#’
return 1+dfs(a-1,b)+dfs(a+1,b)+dfs(a,b+1)+dfs(a,b-1);
//四个方向。又因为七点也是一个，所以加一
}
}
int main()
{
int i,j,p,q,sum;
while(~scanf("%d%d",&w,&h)&&w&&h)
{
for(i=0;i<h;i++)
scanf("%s",str[i]);
for(i=0;i<h;i++)//h行
{
for(j=0;j<w;j++)//w列
{
if(str[i][j]=='@')
p=i,q=j;
}
}
sum=dfs(p,q);
cout << sum <<endl;
}
return 0;
}