Sgu 131-hardwood floor

Question: How many paving methods are there to store the same I-and L-type floor tiles on the M * n floor.

Analysis: DP, combination, and count. Classic DP problems, status compression. Like zoj1100, there are only a few more States.

Status: If f (I, j) is filled with the front I-1, the bit of the I row is represented as the number of paving types when J;

Transfer: I-type bricks, because they can only be horizontal or vertical, there are only two statuses before a brick is paved;

If the current vertical bar is placed, the next row will be affected, and the State correspondence between the two adjacent rows will be established;

L-type brick, because it will certainly affect the two layers, a total of four placement methods, the establishment of adjacent Zeng relationship;

Here we can use DFS to find all the front States F (I-1, k) Addition of the last line of all f (I, j;

F (I, j) = sum (f (I-1, k) {Where F (I-1, k) can generate f (I, j) status };

(The storage status is prepared, but MLE is used, so the last side is extended while the other side is added .)

Note: Only Use % i64d, which is the opposite of zoj. ).

#include <stdio.h>#include <stdlib.h>#include <string.h>long long F[ 10 ][ 1<<9 ];//用dfs找到可以到达本状态的上层的状态 void dfs( int A, int B, int C, int D ){    if ( !A ) {        F[ C ][ D ] += F[ C-1 ][ B ];        return;    }else {        int V = A&-A;//取得最后一个 1的位置         if ( B&V ) {            dfs( A&~V, B&~V, C, D ); //竖着放 1*2             if ( V>1 && (B&(V>>1)) ) dfs( A&~V, B&~(3*(V>>1)), C, D );            if ( B&(V<<1) ) dfs( A&~V, B&~(3*V), C, D );        }         if ( A&(V<<1) ) {            dfs( A&~(3*V), B, C, D );            if ( B&V ) dfs( A&~(3*V), B&~V, C, D );            if ( B&(V<<1) ) dfs( A&~(3*V), B&~(V<<1), C, D );        }    }}int main(){    int n,m;    while ( scanf("%d%d",&n,&m) != EOF && m ) {                if ( m>n ) {int t = m;m = n;n = t;}                int M = (1<<m)-1;                for ( int i = 0 ; i <= n ; ++ i )        for ( int j = 0 ; j <= M ; ++ j )            F[ i ][ j ] = 0LL;        F[ 0 ][ M ] = 1LL;                for ( int i = 1 ; i <= n ; ++ i )        for ( int j = 0 ; j <= M ; ++ j )            dfs( j, M, i, j );                printf("%I64d\n",F[ n ][ M ]);    }    return 0;}

Sgu 131-hardwood floor