HDU 1312 Red and Black(基础bfs或者dfs)

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11843    Accepted Submission(s): 7380

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

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

题目大意：

　　给你一个h*w的棋盘。让你在这个棋盘中找出由@点开始走的‘.’ 有多少个，并且每次只能走相邻的四个方向。

解题思路：

　　这题bfs和dfs都能过的，为的就是加强dfs和bfs的思想，最近做的比赛的时候，感觉自己对于基础算法的理解还不是很到位，所以，加强这方面的练习，从最为基础的

题目来加深理解。bfs的时候，我们要把node压入队列，并且在每次can_move判断后就要对其进行新的入队操作。然后，bfs的解题思路就是说从当前的@点开始，一步一步的走完所有@周围的点，每次我们的扩展按照从“从小到大”的原则来进行，也就是说，我们第一次扩展是把距离@为1的点加入队列，把这个状态结束后，我们就把距离@为2的点加入队列，当这个状态结束后，我们再把距离@为3的点加入队列，，，一直到我们扩展完所有的状态最终使得我们的队列为空，这个时候，我们的解就找到了。

　　而用dfs来做的时候，就相当于一条路走到黑的原则，每找到一个'.'点后，就用'#'来替换他，直到跑完整个地图的所有位置，输出'#'的位置就可以了， 实际上就是can_move()函数的编写了.

bfs代码：

# include<cstdio>
# include<iostream>
# include<queue>
# include<cstring>

using namespace std;

# define MAX 23

char grid[MAX][MAX];
int book[MAX][MAX];
int w,h;

int nxt[4][2] = {{1,0},{0,-1},{-1,0},{0,1}};

struct node
{
    int x,y;
    char val;
};

int can_move ( int x,int y )
{
    if ( x>=0&&x<h&&y>=0&&y<w&&book[x][y]==0&&grid[x][y]=='.' )
        return 1;
    else
        return 0;
}

int bfs ( node start )
{
    int cnt = 0;
    queue<node>Q;
    Q.push(start);
    start.val = '#';
    while ( !Q.empty() )
    {
        node now = Q.front();
        Q.pop();
        for ( int i = 0;i < 4;i++ )
        {
            int tx = now.x+nxt[i][0], ty = now.y+nxt[i][1];
            if ( can_move(tx,ty) )
            {
                book[tx][ty] = 1;
                node newnode;
                newnode.x = tx, newnode.y = ty, newnode.val = '#';
                cnt++;
                Q.push(newnode);
            }
        }
    }
    return cnt;
}


int main(void)
{
    while ( scanf("%d%d",&w,&h)!=EOF )
    {
        if ( w==0&&h==0 )
            break;
        memset(grid,0,sizeof(grid));
        for ( int i = 0;i < h;i++ )
        {
            scanf("%s",grid[i]);
        }
        int stx,sty;
        for ( int i = 0;i < h;i++ )
        {
            for ( int j = 0;j < w;j++ )
            {
                if ( grid[i][j]=='@' )
                {
                    stx = i;
                    sty = j;
                }
            }
        }
        book[stx][sty] = 1;
        node start;
        start.x = stx, start.y = sty, start.val = '@';
        int ans = bfs(start);
        printf("%d
",ans+1);
        memset(book,0,sizeof(book));
    }


    return 0;
}

dfs代码:

# include<cstdio>
# include<iostream>
# include<cstring>

using namespace std;

# define MAX 23

char grid[MAX][MAX];
int book[MAX][MAX];
int nxt[4][2] = {{1,0},{0,-1},{-1,0},{0,1}};
int h,w;
int cnt;

int can_move ( int x,int y )
{
    if( x>=0&&x<h&&y>=0&&y<w&&book[x][y]==0&&grid[x][y]=='.' )
        return 1;
    else
        return 0;
}

void dfs ( int x,int y )
{
    book[x][y] = 1;
    for ( int i = 0;i < 4;i++ )
    {
        int tx = x+nxt[i][0], ty = y+nxt[i][1];
        if ( can_move(tx,ty) )
        {
            cnt++;
            book[tx][ty] = 1;
            dfs(tx,ty);
        }
    }
    return;
}

int main(void)
{
    while ( scanf("%d%d",&w,&h)!=EOF )
    {
        if ( h==0&&w==0 )
            break;
        cnt = 0;
        memset(grid,0,sizeof(grid));
        for ( int i = 0;i < h;i++ )
        {
            scanf("%s",grid[i]);
        }
        int stx,sty;
        for ( int i = 0;i < h;i++ )
        {
            for ( int j = 0;j < w;j++ )
            {
                if ( grid[i][j]=='@' )
                {
                    stx = i;
                    sty = j;
                }
            }
        }
        dfs(stx,sty);
        printf("%d
",cnt+1);
        memset(book,0,sizeof(book));
    }


    return 0;
}