HDU-1312 - BFS与DFS做法

题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=1312

Red and Black Oil Deposits Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 29121 Accepted Submission(s): 17622

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

---

DFS做法：

#include 
using namespace std;

const int MAXN = 21;
char a[MAXN][MAXN];
int ans=0;
int m,n;
int x,y;

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

void dfs(int x,int y)
{
    a[x][y]='#';                        //走过则标记为#
    ans++;
    int p, q;
    for (int i = 0; i < 4; i++)
    {
        p = x + d[i][0];
        q = y + d[i][1];
        if (p > 0 && q > 0 && p <= n && q <= m && a[p][q] == '.')    //对四个方向进行判断是否走过
        {
            dfs(p, q);                   //如果没走过则进行dfs递归
        }
    }
}

int main()
{
    while(cin >> m >> n && m && n)
    {
        ans = 0;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
            {
                cin >> a[i][j];
                if(a[i][j] == '@')
                {
                    x=i;
                    y=j;
                }
            }
            dfs(x,y);
        cout << ans << endl;
    }
    return 0;
}

---

BFS做法：

#include 
using namespace std;
char mp[100][100];
int dir[4][2] = {-1,0,0,-1,1,0,0,1};      //方向
int n,m,ans;

#define CHECK(x, y) (x=0 && y >=0 && y//判断是否越界

struct node                             //把坐标定义成结点
{
    int x,y;
};



void bfs(int dx,int dy)
{
    ans=1;
    queue <node> q;                        //bfs需要用到队列
    node start,next;                       //开两个结点
    start.x = dx;
    start.y = dy;

    q.push(start);                         //把第一个结点放入队列

    while(!q.empty())                       //当队列不为空时
    {
        start = q.front();                  //拿出队列的第一个元素
        q.pop();                            //删除队列里的第一个元素

        for(int i=0; i<4; i++)
        {
            next.x = start.x + dir[i][0];
            next.y = start.y + dir[i][1];
            if( CHECK(next.x,next.y) && mp[next.x][next.y]=='.')
            {
                mp[next.x][next.y] = '#';
                ans++;
                q.push(next);                //把下次走的四个方向放入队列
            }								//依次循环直至走完所以的点
        }
    }
}

int main()
{
    int x,y,dx,dy;
    while(cin >> m >> n && m && n)
    {

        for (y = 0; y < n; y++)
        {
            for (x = 0; x < m; x++)
            {
                cin >> mp[x][y];
                if(mp[x][y] == '@')
                {
                    dx = x;
                    dy = y;
                }
            }
        }
        ans=0;
        bfs(dx,dy);
        cout << ans << endl;
    }
    return 0;
}