Red and Black(BFS or DFS) 分类： dfs bfs 2015-07-05 22:52 2人阅读 评论(0) 收藏

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
",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
",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。

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