Zoj2100--Seeding（Dfs）

Seeding

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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It is spring time and farmers have to plant seeds in the field. Tom has a nice field, which is a rectangle with n * m squares. There are big stones in some of the squares.

Tom has a seeding-machine. At the beginning, the machine lies in the top left corner of the field. After the machine finishes one square, Tom drives it into an adjacent square, and continues seeding. In order to protect the machine, Tom will not drive it into a square that contains stones. It is not allowed to drive the machine into a square that been seeded before, either.

Tom wants to seed all the squares that do not contain stones. Is it possible?

Input

The first line of each test case contains two integers n and m that denote the size of the field. (1 < n, m < 7) The next n lines give the field, each of which contains m characters. 'S' is a square with stones, and '.' is a square without stones.

Input is terminated with two 0's. This case is not to be processed.

Output

For each test case, print "YES" if Tom can make it, or "NO" otherwise.

Sample Input

4 4
.S..
.S..
....
....
4 4
....
...S
....
...S
0 0

Sample Output

YES
NO

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Author: ZHANG, Zheng
Source: Zhejiang University Local Contest 2004

re:拖拉机从左上角开始犁地，黎过的地不能再走回去， 有石头的地不能穿过。看拖拉机能不能走遍所有没有石头的地；

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
char map[10][10]; int ac[4][2]={1, 0, -1, 0, 0, 1, 0, -1};
int m, n, cnt, flag;
void Dfs(int x, int y)
{
    int i, j, nx, ny;                         //注意变量设置位置；
    if(cnt == 0)
    {
        flag = 1;
        return;
    } 
    if(flag)
        return;
    for(i=0; i<4; i++)
    {
        nx = x + ac[i][0];
        ny = y + ac[i][1];
        if(nx >= 1 && nx <=m && ny >= 1 && ny <= n && map[nx][ny] == '.')
        {
            map[nx][ny] = 'S';                    /***********/
            cnt--;
            Dfs(nx, ny);
            cnt++;
            map[nx][ny] = '.';                    /*********/
        } 
    } 
}
int main()
{
    while(~scanf("%d %d", &m, &n), m|n)
    {
        int i, j;  cnt  = n * m;  flag = 0;
        for(i=1; i<=m; i++)
        {
            for(j=1; j<=n; j++)
            {
                cin >> map[i][j];
                if(map[i][j] == 'S')
                    cnt--; 
            }
        }
        cnt--;
        map[1][1] = 'S';
        Dfs(1, 1);
        if(flag)
            printf("YES
");
        else
            printf("NO
");
    }    
    return 0;
}

//找到一个BUG， 爽~~