Red and Black（DFS）

最新推荐文章于 2022-06-06 18:51:49 发布

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本文介绍了一道经典的图遍历问题——在一个由红黑瓷砖组成的矩形房间中，从一个指定的黑色瓷砖出发，计算可以到达的所有黑色瓷砖的数量。文章提供了两种解决方案：深度优先搜索（DFS）和广度优先搜索（BFS），并详细展示了每种方法的实现代码。

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1312

Problem：

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9944 Accepted Submission(s): 6190

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

Source

Asia 2004, Ehime (Japan), Japan Domestic

dfs code：

#include<stdio.h>
#include<string.h>
char map[21][21];
int vis[21][21];
int d[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
int w,h,cnt,sx,sy;
void dfs(int x,int y)
{
int i,xi,yi;
cnt++;
vis[x][y]=1;
for(i=0;i<4;i++)
{
xi=d[i][0]+x;
yi=d[i][1]+y;
if(xi>=0&&xi<h&&yi>=0&&yi<w&&map[xi][yi]=='.'&&!vis[xi][yi])
{

dfs(xi,yi);
}
}
}
int main()
{
int i,j;
while(scanf("%d%d",&w,&h)!=EOF)
{
if(w==0&&h==0)
break;
for(i=0;i<h;i++)
{
scanf("%s",map[i]);
for(j=0;j<w;j++)
{
if(map[i][j]=='@')
{
sx=i;
sy=j;
}
}
}
cnt=0;
memset(vis,0,sizeof(vis));
dfs(sx,sy);
printf("%d\n",cnt);
}
return 0;
}