HDOJ 1312 DFS&BFS

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7337    Accepted Submission(s): 4591

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

45 59 6 13

 

广度搜索，需要用到队列：

//bfs
#include<stdio.h>
int map[100][100];
int queue[10000][2];//队列
int sx[4]={0,1,0,-1};
int sy[4]={1,0,-1,0};
int h,w;
int ans;
void bfs(int x,int y)
{
    int rear=0,front=0;
    queue[rear][0]=x;//enter queue
    queue[rear][1]=y;
    rear++;
    while(front<rear)
    {
        for(int t=0;t<4;t++)
        {
            x=queue[front][0]+sx[t];
            y=queue[front][1]+sy[t];
            if(map[x][y]=='.'&&x>=1&&x<=w&&y>=1&&y<=h)
            {
                queue[rear][0]=x;
                queue[rear][1]=y;
                map[x][y]='#';
                rear++;
                ans++;
            }
        }
        front++;
    }
    
}
int main()
{
    int i,j;
    int x,y;
    while(scanf("%d%d",&h,&w)!=EOF)
    {
        if(!h&&!w)break;
        ans=0;
        for(i=1;i<=w;i++)
        {
            getchar();
            for(j=1;j<=h;j++)
            {
                scanf("%c",&map[i][j]);
                if(map[i][j]=='@')
                    x=i,y=j;
            }
        }
        bfs(x,y);
        printf("%d\n",++ans);
    }
    return 0;
}

深度搜索：

#include<stdio.h>
int map[30][30];
int t,n,m;
int sx[4]={0,0,-1,1};
int sy[4]={-1,1,0,0};
void dfs(int h,int l)//深度搜索
{
    int i,hx,hy;
    t++;
    map[h][l]='@';
    for(i=0;i<4;i++)
    {
        hx=h+sx[i];
        hy=l+sy[i];
        if(hx>=1&&hx<=m&&hy>=1&&hy<=n&&map[hx][hy]=='.')
            dfs(hx,hy);
    }
}
int main()
{
    int a,b,x,y;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)break;
        for(a=1;a<=m;a++)
        {
            getchar();
            for(b=1;b<=n;b++)
            {
                scanf("%c",&map[a][b]);
                if(map[a][b]=='@')
                {
                    x=a;
                    y=b;
                }
            }
        }
        t=0;
        dfs(x,y);
        printf("%d\n",t);
    }
    return 0;
}