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
BFS
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int w,h,sx,sy,cnt;
char map[30][30];
int vis[30][30];
int dx[]={0,0,-1,1};
int dy[]={1,-1,0,0};
struct Node
{
int x;
int y;
}Q[450];
Node s;
void bfs()
{
int front=0,rear=0;
Q[rear++]=s;
while(front<rear)
{
Node t=Q[front++];
for(int i=0;i<4;i++)
{
int x0=t.x+dx[i];
int y0=t.y+dy[i];
Node f;
f.x=x0;f.y=y0;
if(!vis[f.x][f.y]&&f.x>=0&&f.x<h&&f.y>=0&&f.y<w&&map[f.x][f.y]!='#')
{
vis[f.x][f.y]=1;
Q[rear++]=f;
if(map[f.x][f.y]=='.')
cnt++;
}
}
}
}
int main()
{
while(~scanf("%d%d",&w,&h))
{if(w==0||h==0)
break;
memset(vis,0,sizeof(vis));
memset(map,0,sizeof(map));
for(int i=0;i<h;i++)
{
scanf("%s",map[i]);
for(int j=0;j<w;j++)
if(map[i][j]=='@')
{
s.x=i;
s.y=j;
break;
}
}
cnt=0;
bfs();
printf("%d\n",cnt+1);}
return 0;
}
DFS
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int w,h,sx,sy,cnt;
char map[30][30];
int vis[30][30];
int dx[]={0,0,-1,1};
int dy[]={1,-1,0,0};
void dfs(int x,int y)
{
if(map[x][y]=='.')
cnt++;
if(x<0||x>=h||y<0||y>=w||map[x][y]=='#')
return;
for(int i=0;i<4;i++)
{
int x0=x+dx[i];
int y0=y+dy[i];
if(!vis[x0][y0])
{
vis[x0][y0]=1;
dfs(x0,y0);
}
}
}
int main()
{
while(~scanf("%d%d",&w,&h))
{if(w==0||h==0)
break;
memset(vis,0,sizeof(vis));
memset(map,0,sizeof(map));
for(int i=0;i<h;i++)
{
scanf("%s",map[i]);
for(int j=0;j<w;j++)
if(map[i][j]=='@')
{
sx=i;
sy=j;
break;
}
}
cnt=0;
dfs(sx,sy);
printf("%d\n",cnt+1);}
return 0;
}
//6 9
//....#.
//.....#
//......
//......
//......
//......
//......
//#@...#
//.#..#.
可以看出：写BFS时一般要有结构体来表示状态。
求最短路一般用BFS，其他的可能更多用的是DFS
两者的关键都在于找转态。
版权声明：本文为博主原创文章，未经博主允许不得转载。