HDU1312:Red and Black(DFS)
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https://blog.csdn.net/u012325397/article/details/26818317
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.
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)
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
DFS水题,带来了久违一次AC。太感动了~
#include<stdio.h>
#include<string.h>
char map[25][25];
int vis[25][25],cnt,w,h;
void dfs(int x,int y)
{
if(x<0||y<0||x>=w||y>=h)
return ;
if(map[y][x]=='#')
return ;
if(vis[y][x])
return;
if(!vis[y][x]&&(map[y][x]=='.'||map[y][x]=='@'))
{
vis[y][x]=1;
cnt++;
}
dfs(x+1,y);
dfs(x-1,y);
dfs(x,y+1);
dfs(x,y-1);
}
int main()
{
int x,y;
while (scanf("%d%d",&w,&h),w&&h)
{
getchar();
for(int i=0;i<h;++i)
{
for(int j=0;j<w;++j)
{
scanf("%c",&map[i][j]);
if(map[i][j]=='@')
{
x=j;y=i;
}
}
getchar();
}
memset(vis,0,sizeof(vis));
cnt=0;
dfs(x,y);
printf("%d\n",cnt);
}
return 0;
}
posted on 2019-05-01 13:35 xfgnongmin 阅读(117) 评论(0) 编辑 收藏 举报