HDU 4949 light (plug DP, bit operation)

I didn't see the question during the competition. After the competition, I felt that I could be able to answer the question. After the competition, I found that the question was actually a problem ....

Because after writing, there is no limit to tle...

It wasn't until I used bitwise operations to replace the decode, encode, and shift functions commonly used by DP and changed the hash value... 7703 Ms

 

Question: to give a 01 matrix of N * m, you can select a grid for two operations each time. ① flip the neighbor grid ② flip the neighbor grid and yourself. The smallest total number of output operations makes the matrix all 0.

 

Obviously, each grid has four operations (1. No operation; 2. ①; 3. ①; 4. ② ).

At the beginning, two bits are used to indicate a plug. One bits are used to indicate that the current grid is flipped, And the other bits indicate that the source of the plug needs to be flipped. Then 2*3*(4 ^ 10) is not scientific.

Later we found that, in fact, we can classify it like this: ① don't operate (cost 0); ② flip ourselves (cost 2); ③ flip neighbors (cost 1 ); this is 2*3*(3 ^ 10)

③ Includes two cases. In fact, if you perform the ③ operation on a grid A, the neighbor grid bcde of the grid can turn off the light.

In addition, the answer must be less than twice the number of 1 equal to the first matrix. You can use this method for a certain degree of pruning.

Then there are various bitwise operations .... I am dizzy...

 

In addition, in fact, this plug-in DP consumes a lot of extra costs (clearing the hash table or something ), therefore, the speed is slower than that of direct DP [I] [J] [k ?).

  1 #include <cstdio>  2 #include <cstring>  3 #include <iostream>  4 using namespace std;  5   6   7 #define HASH 100007  8 #define STATE 500010  9 #define maxd 15 10  11 int maze[maxd][maxd]; 12 int code[maxd]; 13 int n,m; 14 struct HASHMAP{ 15     int head[HASH]; 16     int state[STATE],nxt[STATE]; 17     int f[STATE]; 18     int sz; 19     void clear(){sz=0;memset(head,-1,sizeof(head));} 20     void push(int st,int ans){ 21         int h=st%HASH; 22         for(int i=head[h];i!=-1;i=nxt[i]){ 23             if(st==state[i]){ 24                 f[i] = f[i]<ans?f[i]:ans; 25                 return ; 26             } 27         } 28         state[sz]=st,nxt[sz]=head[h],f[sz]=ans; 29         head[h]=sz++; 30     } 31 }hm[2]; 32 void decode(int st){ 33     for(int i=m;i>=0;--i) code[i]=st&3,st>>=2; 34 } 35 int encode(){ 36     int ret=0; 37     for(int i=0;i<=m;++i) ret=ret<<2|code[i]; 38     return ret; 39 } 40 void shift(){ 41     for(int i=m;i;--i) code[i]=code[i-1]; 42     code[0]=0; 43 } 44 int ans; 45 int zo,oz,oo; 46 void dpblank(int i,int j,int cur){ 47     int mv = j==m?2:0; 48     int all = (1<<(2*(m+1)-mv) ) -1; 49     for(int k=0;kView code