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链接:
http://poj.org/problem?id=1321
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 28899 | Accepted: 14307 |
Description
Input
Output
Sample Input
2 1 #. .# 4 4 ...# ..#. .#.. #... -1 -1
Sample Output
2 1
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include <queue> using namespace std; #define N 11 int n, k, vis[N], z, num; char s[N][N]; void DFS(int m) { if(num==k) { z++; return ; } if(m>=n) return ; for(int i=0; i<n; i++) { if(!vis[i] && s[m][i]==‘#‘) { vis[i] = 1; num++; DFS(m+1); vis[i] = 0; num--; } } DFS(m+1); } int main() { while(scanf("%d%d", &n, &k)!=EOF) { if(n==-1 && k==-1) break; int i; memset(s, 0, sizeof(s)); memset(vis, 0, sizeof(vis)); for(i=0; i<n; i++) scanf("%s", s[i]); num = z = 0; DFS(0); printf("%d\n", z); } return 0; }
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原文地址:http://www.cnblogs.com/YY56/p/4782758.html