Poj 2411 Mondriaan's Dream (State compression DP)

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Question:

There are two kinds of dominoes: 1*2 and 2*1. How many paving methods can be used in a rectangle filled with M * n?

Ideas:

In binary format, two consecutive 00 indicates that the two grids are horizontal, and 1 indicates that the grids are vertical (the upper part or the lower part of the bone card ).

First, process and obtain all the statuses that meet the condition. If there is 0, two zeros must appear consecutively.

For the state J of row I, where J is 1, then the same place of the I-1 line must also be 1 or 0. Line I and I-1 are both 1 place, they constitute a vertical card, then these places of I-2 line, either 0 or 1, so we need to change 1 to 0 in recursion, so that the place where the I-2 row can be 1 or 0.

F [I] [J] indicates the status of J in line I.

 

# Include <iostream> # include <cstdio> # include <cstring> # include <cmath> # include <algorithm> # include <vector> using namespace STD; typedef long int64; const int INF = 0x3f3f3f; const int max_state = (1 <11) + 10; const int maxn = 12; int n, m; int sta [max_state], idx; int64 f [maxn] [max_state]; int maxstate; bool OK [max_state]; bool check (INT Sta) {int I = 0; while (I <m) {If (STA> I) & 1) ++ I; else {if (I = S-1 | (S Ta> (I + 1) & 1 )! = 0) return false; I + = 2;} return true;} void Init () {memset (OK, 0, sizeof (OK )); maxstate = (1 <m)-1; idx = 0; For (INT I = 0; I <= maxstate; ++ I) {If (check (I )) {sta [idx ++] = I; OK [I] = true ;}}// int64 DFS (INT cur, int st) in the memory search) {If (F [cur] [st]! =-1) return f [cur] [st]; If (cur = 0) Return OK [st]; int64 ret = 0; For (INT I = 0; I <idx; ++ I) {If (STA [I] & St) = sT) {RET + = DFS (cur-1, st ^ sta [I]) ;}} return f [cur] [st] = ret;} // recursive dpint64 dp () {If (n = 1) return OK [0]; memset (F, 0, sizeof (f); int ans = 0; For (INT I = 0; I <n; ++ I) {if (I = 0) {for (Int J = 0; j <idx; ++ J) f [I] [sta [J] = 1 ;} else {for (int s = 0; S <= maxstate; ++ s) {for (int K = 0; k <idx; ++ K) {If ((STA [k] & S) = s) f [I] [s] + = f [I-1] [sta [k] ^ s] ;}}} for (INT I = 0; I <idx; ++ I) ans + = f [n-1] [sta [I]; return f [n-1] [0];} int main () {While (~ Scanf ("% d", & N, & M) & N + M) {Init (); memset (F,-1, sizeof (f )); // cout <DFS (n-1, 0) <Endl; cout <dp () <Endl ;}return 0 ;}