zoj 2100 Seeding

Seeding

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

#include <iostream>
#include <fstream>
using namespace std;

bool visit[7][7];      // preserve wether the square[i][j] has been visited
int row,col;           // count the row and col of square
int dir[8] = {0,1,0,-1,1,0,-1,0};

int DFS(int r, int c, int times){
if( 0 == times )
return 1;
visit[r][c]=1;
//cout<<"["<<times<<"]: "<<"visit["<<r<<"]["<<c<<"]: "<<visit[r][c]<<'\n';
for(int i = 0 ; i < 8; i+=2){
int rr = r+dir[i];
int cc = c+dir[i+1];
if( rr >= 0 && rr < row && cc >= 0 && cc < col&& visit[rr][cc] != 1 && DFS(rr,cc,times-1))
return 1 ;
}
visit[r][c]=0;
return 0;
}

int main(){
int a;
ifstream fin;
fin.open("zoj2100.txt");
if(!fin)
cerr<<"error!\n";
while(fin>>row>>col){
if(0==row&&0==col)
break;
// initialize the sqare
int count = 0;
char ch;
for(int i = 0; i < row; ++i){
for(int j = 0 ; j < col; ++j){
fin>>ch;
if('.'==ch){
++count;
visit[i][j] = 0;
}else
visit[i][j] = 1;
}
}

if(DFS(0,0,count-1))
cout<<"YES\n";
else
cout<<"NO\n";
}
fin.close();
system("PAUSE");

}

就是简单的DFS，条件判断清楚就好，然后剩下就是剪枝+回溯，结果就出来了。另外值得注意的是，对于输入的square直接进行判断，将S的地方直接表示为visit过得地方，这样在DFS过程中就不需要再判断一次是否是S。

代码还可以写的简练一些，懒得改了