标签:
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
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 }
CodeForcesGym 100212E Long Dominoes
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原文地址:http://www.cnblogs.com/crackpotisback/p/4857717.html