题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17773    Accepted Submission(s):
10826

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
45
59
6
13
题目大意: “.”代表黑地板,“#”代表红地板,“@”代表人的位置,人可以经过黑地板不能经过红地板,问在给定的图中人一共可以经过多少块地板
解题思路:深搜,在每个位置都搜索该位置的上下左右四个位置,将图中能够经过的点全部搜一遍
AC代码:
 #include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
using namespace std;
int w,h;
int a[][] = {{-,},{,},{,-},{,}}; //位置数组
char str[][];
int f[][]; //标记该位置是否走过
int sum;
void dfs(int x,int y)
{
f[x][y] = ;
for (int i = ; i < ; i ++)
{
int x1=x + a[i][];
int y1=y + a[i][];
if (x1 >= && x1 < h && y1 >= && y1 < w && str[x1][y1]!='#' && f[x1][y1] == )
{ //判断边界、是否是不能经过的红地板、该地板是否已经经过
sum ++;
dfs(x1,y1);
}
}
}
int main ()
{
int i,j,x,y;
while (scanf("%d%d",&w,&h),w&&h)
{
for (i = ; i < h; i ++)
scanf("%s",str[i]); memset(f,,sizeof(f));
for (i = ; i < h; i ++)
for (j = ; j < w; j ++)
if (str[i][j] == '@') //找出人的位置
{
x = i;
y = j;
break;
}
sum = ;
dfs(x,y);
printf("%d\n",sum);
}
return ;
}

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