【HDU】5766 Filling【轮廓线dp+burnside引理】

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题目链接：Filling
轮廓线dp出整个的，一半的，1/4的，然后用burnside引理去重。
#include <bits/stdc++.h>using namespace std ;typedef long long LL ;typedef pair < int , int > pii ;#define clr( a , x ) memset ( a , x , sizeof a )const int mod = 1e9 + 7 ;const LL v2 = 500000004 ;const LL v4 = 250000002 ;int dp[2][1 << 21] , n ;int ans[22] = { 0 , 1 , 2 , 2 , 12 , 84 , 1641 , 50788 , 3183091 , 338785462 , 429880385 , 948920428 , 392930107 , 842384602 , 36468923 , 534250025 , 693507546 , 311288183 , 320259213 , 232751393 , 866290697 };;void up ( int& x , int y ) {    x += y ;    if ( x >= mod ) x -= mod ;}void show ( int s ) {    printf ( "s = " ) ;    for ( int i = 0 ; i < n + 1 ; ++ i ) {        printf ( "%d" , s >> i & 1 ) ;    }    puts ( "" ) ;}void solve () {    if ( n == 1 ) {        printf ( "%d\n" , 1 ) ;        return ;    }    int cur = 0 ;    int tot = 1 << n + 1 , x = 1 << n , y = ( 3 << n - 1 ) + 1 ;    clr ( dp , 0 ) ;    dp[cur][0] = 1 ;    for ( int i = 1 ; i < n ; ++ i ) {        for ( int j = 0 ; j < n ; ++ j ) {            cur ^= 1 ;            clr ( dp[cur] , 0 ) ;            for ( int s = 0 ; s < tot ; ++ s ) if ( dp[cur ^ 1][s] ) {                if ( j && !( s & 3 ) && !( s & x ) ) up ( dp[cur][s >> 1 | y] , dp[cur ^ 1][s] ) ;                up ( dp[cur][s >> 1] , dp[cur ^ 1][s] ) ;            }        }    }    int ans0 = 0 , ans180 = 0 , ans90 = 0 , ans90_2 = 0 ;    for ( int s = 0 ; s < tot ; ++ s ) {        up ( ans0 , dp[cur][s] ) ;    }    clr ( dp , 0 ) ;    dp[cur][0] = 1 ;    for ( int i = 1 ; i < n / 2 ; ++ i ) {        for ( int j = 0 ; j < n ; ++ j ) {            cur ^= 1 ;            clr ( dp[cur] , 0 ) ;            for ( int s = 0 ; s < tot ; ++ s ) if ( dp[cur ^ 1][s] ) {                if ( j && !( s & 3 ) && !( s & x ) ) up ( dp[cur][s >> 1 | y] , dp[cur ^ 1][s] ) ;                up ( dp[cur][s >> 1] , dp[cur ^ 1][s] ) ;            }        }    }    if ( n & 1 ) {        for ( int j = 0 ; j < n ; ++ j ) {            cur ^= 1 ;            clr ( dp[cur] , 0 ) ;            for ( int s = 0 ; s < tot ; ++ s ) if ( dp[cur ^ 1][s] ) {                if ( j && !( s & 3 ) && !( s & x ) ) up ( dp[cur][s >> 1 | y] , dp[cur ^ 1][s] ) ;                up ( dp[cur][s >> 1] , dp[cur ^ 1][s] ) ;            }        }    } else {        for ( int j = 0 ; j < n / 2 ; ++ j ) {            cur ^= 1 ;            clr ( dp[cur] , 0 ) ;            for ( int s = 0 ; s < tot ; ++ s ) if ( dp[cur ^ 1][s] ) {                if ( j && !( s & 3 ) && !( s & x ) ) up ( dp[cur][s >> 1 | y] , dp[cur ^ 1][s] ) ;                up ( dp[cur][s >> 1] , dp[cur ^ 1][s] ) ;            }        }    }    for ( int s = 0 ; s < tot ; ++ s ) {        int ok = 1 ;        for ( int j = 1 ; j <= ( n + 1 ) / 2 ; ++ j ) {            if ( ( s & ( 1 << j ) ) && ( s & ( 1 << n + 1 - j ) ) ) {                ok = 0 ;                break ;            }        }        if ( ok ) {            up ( ans180 , dp[cur][s] ) ;            if ( n % 2 == 0 && !( s & ( 1 << n ) ) && !( s & 3 ) ) {                up ( ans180 , dp[cur][s] ) ;            }            //if ( dp[cur][s] ) show ( s ) ;        }    }    int m = ( n + 1 ) / 2 ;    tot = 1 << m + 1 ;    x = 1 << m ;    y = 3 << ( m - 1 ) | 1 ;    for ( int o = 0 ; o < 1 << m ; ++ o ) {        if ( ( n & 1 ) && ( o & ( 1 << ( n / 2 ) ) ) ) continue ;        for ( int s = 0 ; s < tot ; ++ s ) {            dp[cur][s] = 0 ;        }        dp[cur][0] = 1 ;        for ( int i = 1 ; i < n / 2 + 1 ; ++ i ) {            for ( int j = 0 ; j < m ; ++ j ) {                cur ^= 1 ;                for ( int s = 0 ; s < tot ; ++ s ) {                    dp[cur][s] = 0 ;                }                for ( int s = 0 ; s < tot ; ++ s ) if ( dp[cur ^ 1][s] ) {                    if ( j && !( s & 3 ) && !( s & x ) ) up ( dp[cur][s >> 1 | y] , dp[cur ^ 1][s] ) ;                    up ( dp[cur][s >> 1] , dp[cur ^ 1][s] ) ;                }            }            for ( int s = 0 ; s < tot ; ++ s ) {                if ( ( s & 1 ) != ( ( o >> i - 1 ) & 1 ) ) dp[cur][s] = 0 ;            }        }        for ( int s = 0 ; s < tot ; ++ s ) {            int ok = 1 ;            for ( int k = 0 ; k < m ; ++ k ) {                if ( ( o >> k ) & ( s >> k + 1 ) ) {                    ok = 0 ;                    break ;                }            }            if ( ok ) {                up ( ans90 , dp[cur][s] ) ;                if ( n % 2 == 0 && !( s & ( 1 << m ) ) && !( s & 1 ) ) {                    up ( ans90 , dp[cur][s] ) ;                }            }        }    }    ans[n] = ( ( ans0 - ans180 + mod ) % mod * v4 + ( ans180 - ans90 + mod ) * v2 + ans90 ) % mod ;    printf ( " , %d" , ans[n] ) ;}int main () {    /*    printf ( "{ 1" ) ;    for ( n = 2 ; n <= 20 ; ++ n ) {        solve () ;    }    printf ( " };\n" ) ;    */    int T ;    scanf ( "%d" , &T ) ;    for ( int i = 1 ; i <= T ; ++ i ) {        scanf ( "%d" , &n ) ;        printf ( "Case #%d: %d\n" , i , ans[n] ) ;    }    return 0 ;}
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