PKU 2411 Mondriaan's dream DP status Compression

 

I have never done any State compression questions before. Today I have seen it. I have read the following brilliant words before .-_-

 

It should have been very early that the question of AC has been adjusted to the present, why? This is why I am in poor state!

 

I read a report in the afternoon. Others used the DFS method. One of them had the same idea as I did, but I didn't have any data. The idea was right at the beginning, the mistake was made under a very small condition. It hurt me. I thought my DP was wrong! It seems that you have to trust your own strength.

 

I think the algorithm is slow and simple. It is no different from the idea of AC in the morning, that is, the binary status of each line.

 

If DP [I] [k] is set to K. Then find out the relationship between it and the state of the I-1 line.

 

Since the I-row DP must fully overwrite the I-1-th row, you only need to grasp this condition and not put it.

 

1. If row I has 0, the row I-1 must be 1;

 

2. If column X in row I is column 1 and row I is 0, the row I must be vertical;

 

3. If column X in row I is 1 and the column in row I-1 is also 1, there is only one possibility, row I is horizontal, therefore, the column x + 1 of row I must also be 1, and because column x + 1 of row I is 1, so the x + 1 column of the I-1 row must also be 1.

 

It is easy to find the link. Oh, I was upset when I thought of it. I should have hated AC for a long time.

 

Oh ~, I almost forgot, there is also the initialization of the second line, initialize it to exist only horizontally or not.

 

This paragraph from: http://gisyhy.blog.163.com/blog/static/129390343200992441558735/

 

PS: the original author's code is beautifully written.

 

 

# Include <stdio. h> <br/> # include <string. h> <br/> # include <stdlib. h> <br/> # include <math. h> <br/> # include <iostream> <br/> # include <string> <br/> # include <vector> <br/> # include <queue> <br /># include <deque> <br/> # include <list> <br/> # include <algorithm> <br/> # include <map> <br/> # include <pash_map> <br/> # include <pash_set> <br/> # include <set> <br/> # include <utility> <br/> using namespace STD; <br/> long DP [11] [1 <11]; <br/> int H, W; <br/> bool isok (int s) <br/>{< br/> int I = 0; <br/> while (I <W) <br/> {<br/> If (S & (1 <I )) <br/> {<br/> if (I = W-1 | (S & (1 <(I + 1) = 0) <br/> return false; <br/> I + = 2; <br/>}< br/> else <br/> I ++; <br/>}< br/> return true; <br/>}< br/> bool isok (INT S1, int S2) <br/>{< br/> int I = 0; <br/> while (I <W) <br/> {<br/> If (S1 & (1 <I )) <br/> {<br/> If (s2 & (1 <I) = 0) <br/> I ++; <br/> else <br/> {<br/> if (I = W-1 | (S1 & (1 <(I + 1 ))) & (s2 & (1 <(I + 1) = 0) <br/> return false; <br/> I + = 2; <br/>}< br/> else <br/> {<br/> If (s2 & (1 <I) = 0) <br/> return false; <br/> I ++; <br/>}< br/> return true; <br/>}< br/> int main () <br/> {<br/> while (scanf ("% d", & H, & W ), H & W) <br/>{< br/> If (H <W) <br/>{< br/> int TMP = W; <br/> W = H; <br/> H = TMP; <br/>}< br/> memset (DP, 0, sizeof (DP )); <br/> int max = (1 <W); <br/> for (INT I = 0; I <Max; ++ I) <br/> If (isok (I) <br/> DP [0] [I] = 1; <br/> for (INT I = 1; I <p; + + I) <br/> for (Int J = 0; j <Max; ++ J) // line I <br/> for (int K = 0; k <Max; ++ K) // line I-1 <br/> If (isok (j, k )) <br/> DP [I] [J] + = DP [I-1] [k]; <br/> printf ("% i64d/N ", DP [h-1] [MAX-1]); <br/>}< br/> return 0; <br/>}