Red and Black(BFS or DFS)

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。