棋盘问题

标签：

2 1
#.
.#
4 4
...#
..#.
.#..
#...
-1 -1

2
1

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define LL long long
#define pii pair<string, int>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
using namespace std;
char mmap[10][10];
bool vis[10];
int n,k;
int dfs(int row,int coun){
   //cout<<row<<' '<<coun<<"**********"<<endl;
   int sum=0;
   if(coun == k) return 1;
   if(row < n){
       for(int i=0;i<n;i++){
         if(vis[i]==false&&mmap[row][i] == '#'){
           vis[i] = true;
           sum+=dfs(row+1,coun+1);
           vis[i] = false;
         }
       }
       sum+=dfs(row+1,coun);
   }
   return sum;
}
int main (){
  while(~rd2(n,k)&&n!=-1&&k!=-1){
        memset(vis,0,sizeof(vis));
        for(int i=0;i<n;i++){
            scanf("%s",mmap[i]);
        }
        printf("%d\n",dfs(0,0));
  }
return 0;
}

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原文地址：http://blog.csdn.net/u014665013/article/details/51328829