题目链接：HDU1312 / POJ1979 / ZOJ2165 Red and Black（红与黑）
Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9902 Accepted Submission(s): 6158
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
Source
Asia 2004, Ehime (Japan), Japan Domestic
题意：
有一个长方形的房间布满了正方形的瓷砖，瓷砖要么红色要么黑色。一男子站在其中一块黑色瓷砖上，可向上下左右四个方向移动，但不能移动到红色瓷砖上，问他可到达的黑色瓷砖数量。
分析：
DFS搜索。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dir[4][2] = {{-1, 0},{0, 1},{1, 0},{0, -1}};
int cnt, W, H;
char mp[21][21];
bool vis[21][21];
void dfs(int x, int y)
{
vis[x][y] = true;
for(int i = 0; i < 4; i++)
{
int tx = x + dir[i][0];
int ty = y + dir[i][1];
if(tx >= 1 && tx <= H && ty >= 1 && ty <= W && !vis[tx][ty] && mp[tx][ty] == '.')
{
cnt++;
dfs(tx, ty);
}
}
}
int main()
{
char c;
int x, y;
while(scanf("%d%d", &W, &H), W, H)
{
scanf("%c", &c);
for(int i = 1; i <= H; i++)
{
for(int j = 1; j <= W; j++)
{
scanf("%c", &mp[i][j]);
if(mp[i][j] == '@')
{
x = i;
y = j;
}
}
scanf("%c", &c);
}
cnt = 1;
memset(vis, false, sizeof(vis));
dfs(x, y);
printf("%d\n", cnt);
}
return 0;
}