Floor Tile | Pressure DP Exercises

There is a n*m (n<=5,m<=1000) of the chessboard, now there are 1*2 and 2*1 of small pieces of wood, to cover the entire board, how many ways? The answer only needs mod1,000,000,007.

I don't know the source of the problem, Qaq.

The range of N and M should be the same, but the value of n given by the topic is very small, which gives us the idea of using the pressure DP.

Assuming that the first column is already full, the second column is only related to the first column's effect on it, and similarly, the third column is only related to the second column's effect on it, and we can use the binary to represent the condition of a given column, which represents the state of a column, such as State=4, which has a column status of 00100. Use dp[i][state] to denote column I, the effect of column i-1 on it is the number of scenarios for state, and the number of scenarios per column can be achieved by searching, Dp[i][state]=sigma (Dp[i-1][la]) La can be filled into state.

Code:

1 //Paving Board2 //2015/10/223#include <cstdio>4#include <iostream>5#include <cstdlib>6#include <cmath>7#include <vector>8#include <cstring>9#include <algorithm>Ten #defineMAXN 100000000+50 One #defineINF 0x7fffffff A #defineXiao 1e-9 - #defineMoD 1000000007 - using namespacestd; the intdp[1005][ +],n,m; - voidDfsintIintJintStateintnext) - { -     if(j==N) +     { -dp[i+1][next]+=Dp[i][state]; +dp[i+1][next]%=MoD; A         return; at}//If the last row is enumerated, then the next column status is next, plus the number of scenarios for which the column status is State -     if(((1&LT;&LT;J) &state) >0) DFS (i,j+1, State,next);//if the line J position is already occupied, skip directly, search j+1 line -     if(((1&LT;&LT;J) &state) = =0) DFS (i,j+1, state,next| (1&LT;&LT;J));//if not occupied, try to fill in a 1*2 -     if(j+1<n&& (1&LT;&LT;J) &state) = =0&& (1<< (j+1) &state) = =0)) DFS (i,j+2, State,next);//If this location and the next position are not occupied, try putting a 2*1 -     return; - } in intMain () - { toCin>>n>>m; +Memset (DP,0,sizeof(DP)); -dp[1][0]=1; the      for(intI=1; i<=m;++i) *        for(intj=0;j< (1<<n); + +j) $         {Panax Notoginseng             if(Dp[i][j]) DFS (i,0J0); -                     } thecout<<dp[m+1][0]<<Endl; +     return 0; A}

Floor Tile | Pressure DP Exercises