Zoj 1100 Paving Brick State compression

1, the ST is stored in each row of all possible, 0 is empty, 1 is occupied, st[i][0] in the possible state of the deposit Line (From,st[i][1] in the next line of the state that matches with to;

2, Dfs from is a row of the first n lattice state, to is the state of the match;

3, because only look at the possible state of each row, so DFS in the n==w when the possibility has been taken, exit recursion;

4, for the row of the nth column, take three kinds of the way of placing rectangles

DFS (n+2, (from<<2) +3, (to<<2) +3); Horizontal, this layer and the next layer match more than two bits 1 (binary 11 is 3)
DFS (n+1, (from<<1) +1,to<<1);//vertical, this layer one more 1, matching the lower level this position is 0
DFS (n+1,from<<1, (to<<1) +1);//Do not put, this layer more than a 0, matching the lower layer of more than 1

5. Dp[i][j] The number of methods in which the state of deposit I is placed at J, the boundary is to be fully filled, dp[0][(1<<w) -1]=1, the last requirement is dp[h][(1<<W)-1];

6. DP process is

for (i=1;i<=h;i++)
{
for (j=0;j<cnt;j++)
{
Dp[i][st[j][1]]+=dp[i-1][st[j][0]];
}

}

#include <stdio.h> #include <string.h> int st[3000][2]; int dp[11][3000];
2^11=2048 Maximum number of States int cnt,h,w;
    void Dfs (int n,int from,int to)//enumerates all rows of possible and matching underlying {if (n>w) return;
        if (n==w)//This line is just finished, get a row of combinations {st[cnt][0]=from;
        St[cnt][1]=to;
        cnt++;
    Return } dfs (n+2, (from<<2) +3, (to<<2) +3); Horizontal, this layer and the next layer match more than two bit 1 Dfs (n+1, (from<<1) +1,to<<1);//vertical, this layer one more 1, matching the lower level this position is 0 Dfs (n+1,from<<1, (to
    <<1) +1);//Do not put, this layer more than one 0, matching the underlying more than 1} int main () {int i,j;
            while (scanf ("%d%d", &h,&w) &&h) {if ((w*h)%2) {printf ("0\n");
        Continue
        } memset (Dp,0,sizeof (DP));
        memset (st,0,sizeof (ST)); cnt=0;
        Record the total number of possible states in each row DFS (0,0,0);
                dp[0][(1<<w) -1]=1;//boundary for (i=1;i<=h;i++) {for (j=0;j<cnt;j++) {
    Dp[i][st[j][1]]+=dp[i-1][st[j][0]];        }} printf ("%d\n", dp[h][(1&LT;&LT;W)-1]);
} return 0;
 }