USACO 5.4 All Latin Squares(DFS)

打表水过。看的题解，置换的那个优化完全没有看懂。

 1 /*
 2 ID: cuizhe
 3 LANG: C++
 4 TASK: latin
 5 */
 6 #include <cstdio>
 7 #include <iostream>
 8 using namespace std;
 9 int r[11][11],c[11][11];
10 int ans,n;
11 void dfs(int x,int y)
12 {
13     int xx,yy,i;
14     if(x > n-1)
15     {
16         ans ++;
17         return ;
18     }
19     if(y+1 > n)
20     {
21         xx = x+1;
22         yy = 2;
23     }
24     else
25     {
26         xx = x;
27         yy = y+1;
28     }
29     for(i = 1;i <= n;i ++)
30     {
31         if(!r[x][i]&&!c[y][i])
32         {
33             r[x][i] = 1;
34             c[y][i] = 1;
35             dfs(xx,yy);
36             r[x][i] = 0;
37             c[y][i] = 0;
38         }
39     }
40 }
41 int main()
42 {
43     int cl[10],i;
44     freopen("latin.in","r",stdin);
45     freopen("latin.out","w",stdout);
46     cl[0] = 1;
47     for(i = 1;i <= 7;i ++)
48     cl[i] = i*cl[i-1];
49     scanf("%d",&n);
50     if(n == 7)
51     {
52         printf("12198297600\n");
53         return 0;
54     }
55     for(i = 1;i <= n;i ++)
56     {
57         r[i][i] = 1;
58         c[i][i] = 1;
59     }
60     dfs(2,2);
61     printf("%d\n",cl[n-1]*ans);
62     return 0;
63 }

 

posted @ 2013-05-28 08:46  Naix_x  阅读(447)  评论(0)  编辑  收藏  举报