CodeForcesGym 100212E Long Dominoes

Long Dominoes

Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on CodeForcesGym. Original ID: 100212E
64-bit integer IO format: %I64d      Java class name: (Any)

Find the number of ways to tile an m*n rectangle with long dominoes -- 3*1 rectangles.

Each domino must be completely within the rectangle, dominoes must not overlap (of course, they may touch each other), each point of the rectangle must be covered.


Input

The input contains several cases. Each case stands two integers m and n (1 <= m <= 9, 1 <= n <= 30) in a single line. The input ends up with a case of m = n = 0.


Output

Output the number of ways to tile an m*n rectangle with long dominoes.


Sample Input

3 3
3 10
0 0

Sample Output

2
28

 

Source

Andrew Stankevich's Contest #4

Author

Andrew Stankevich
 
解题：状压dp
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 1<<18;
 5 LL dp[2][maxn];
 6 vector<int>g[maxn];
 7 bool tab[10][10];
 8 int stx[maxn],tot;
 9 void dfs(int row,int st,int n) {
10     if(row == n) {
11         int tst = 0;
12         for(int i = n-1; i >= 0; --i) {
13             tst <<= 2;
14             tst |= tab[i][1]|(tab[i][2]<<1);
15         }
16         g[tst].push_back(st);
17         stx[tot++] = tst;
18         stx[tot++] = st;
19         return;
20     }
21     if(!tab[row][0]) {
22         if(!tab[row][1] && !tab[row][2]) {
23             tab[row][0] = tab[row][1] = tab[row][2] = true;
24             dfs(row + 1,st,n);
25             tab[row][0] = tab[row][1] = tab[row][2] = false;
26         }
27         if(row + 3 > n || tab[row + 1][0] || tab[row + 2][0]) return;
28         tab[row + 2][0] = tab[row + 1][0] = tab[row][0] = true;
29         dfs(row + 3,st,n);
30         tab[row + 2][0] = tab[row + 1][0] = tab[row][0] = false;
31     } else dfs(row + 1,st,n);
32 }
33 void init(int st,int n) {
34     memset(tab,false,sizeof tab);
35     for(int i = 0,xst = st; i < n; ++i,xst >>= 2) {
36         int row = xst&3;
37         tab[i][0] = row&1;
38         tab[i][1] = (row>>1)&1;
39         if(row == 2) return;
40     }
41     dfs(0,st,n);
42 }
43 int main() {
44     freopen("dominoes.in","r",stdin);
45     freopen("dominoes.out","w",stdout);
46     int m,n;
47     scanf("%d%d",&m,&n);
48     for(int i = 0; i < (1<<(m + m)); ++i) init(i,m);
49     sort(stx,stx + tot);
50     tot = unique(stx,stx + tot) - stx;
51     int cur = dp[0][0] = 1;
52     for(int i = 1; i <= n; ++i) {
53         for(int j = 0; j < tot; ++j) {
54             for(int k = g[stx[j]].size()-1; k >= 0; --k)
55                 dp[cur][stx[j]] += dp[cur^1][g[stx[j]][k]];
56         }
57         cur ^= 1;
58         memset(dp[cur],0,sizeof dp[cur]);
59     }
60     printf("%I64d\n",dp[cur^1][0]);
61     return 0;
62 }
View Code

 

posted @ 2015-10-06 20:49  狂徒归来  阅读(339)  评论(0)  编辑  收藏  举报