狀壓與矩陣加速的藕斷絲連

• Quad Tiling
• • description
  • solution
  • code
• [Hnoi2010]Bus 公交線路
• • description
  • solution
  • code

Quad Tiling

description

solution

設$d{p}_{i,S}:$ $i$列的狀態為$S$的方案數，最后答案為$d{p}_{n,(1<<4)?1}$（最后一列剛好鋪滿）

• $$?\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}?\to ?\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}?/?\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}?/?\begin{array}{c}0\\ 0\\ 1\\ 1\end{array}?/?\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}?/?\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}?$$

• $$?\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}?\to ?\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}?$$

• $$?\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}?\to ?\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}?/?\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}?$$

• $$?\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}?\to ?\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}?$$

• $$?\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}?\to ?\begin{array}{c}0\\ 0\\ 1\\ 1\end{array}?/?\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}?$$

• $$?\begin{array}{c}0\\ 0\\ 1\\ 1\end{array}?\to ?\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}?/?\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}?$$

• $1010\to 0101...$等的其余狀態都是不可能最后剛好$n$被多米諾骨牌鋪滿而不凸出去幾格的

code

#include <cstdio> #include <cstring> #define int long long int n, mod; struct matrix {int n, m;int c[16][16];matrix() {memset( c, 0, sizeof( c ) );}matrix operator * ( matrix &t ) {matrix ans;ans.n = n, ans.m = t.m;for( int i = 0;i < n;i ++ )for( int j = 0;j < t.m;j ++ )for( int k = 0;k < m;k ++ )ans.c[i][j] = ( ans.c[i][j] + c[i][k] * t.c[k][j] ) % mod;return ans;} }g, ret; int id[2][2][2][2];matrix qkpow( matrix x, int y ) {matrix ans;ans.n = ans.m = x.m;for( int i = 0;i < ans.n;i ++ )ans.c[i][i] = 1;while( y ) {if( y & 1 ) ans = ans * x;x = x * x;y >>= 1;}return ans; }void rebuild() {memset( g.c, 0, sizeof( g.c ) );memset( ret.c, 0, sizeof( ret.c ) );g.n = g.m = ret.m = 1 << 4, ret.n = 1;g.c[id[0][0][0][0]][id[1][1][0][0]] = 1;g.c[id[0][0][0][0]][id[0][0][1][1]] = 1;g.c[id[0][0][0][0]][id[1][0][0][1]] = 1;g.c[id[0][0][0][0]][id[1][1][1][1]] = 1;g.c[id[0][0][0][0]][id[0][0][0][0]] = 1;g.c[id[0][0][1][1]][id[1][1][0][0]] = 1;g.c[id[0][0][1][1]][id[0][0][0][0]] = 1;g.c[id[1][1][0][0]][id[0][0][0][0]] = 1;g.c[id[1][1][0][0]][id[0][0][1][1]] = 1;g.c[id[1][0][0][1]][id[0][0][0][0]] = 1;g.c[id[1][0][0][1]][id[0][1][1][0]] = 1;g.c[id[0][1][1][0]][id[1][0][0][1]] = 1;g.c[id[1][1][1][1]][id[0][0][0][0]] = 1;ret.c[0][id[0][0][0][0]] = 1; }signed main() {for( int i = 0;i < 2;i ++ )for( int j = 0;j < 2;j ++ )for( int k = 0;k < 2;k ++ )for( int w = 0;w < 2;w ++ )id[i][j][k][w] = w + ( 1 << 1 ) * k + ( 1 << 2 ) * j + ( 1 << 3 ) * i;while( scanf( "%lld %lld", &n, &mod ) ) {if( ! n && ! mod ) return 0;rebuild();g = qkpow( g, n );ret = ret * g;printf( "%lld\n", ret.c[0][id[0][0][0][0]] );}return 0; }

[Hnoi2010]Bus 公交線路

description

solution

設$d{p}_{i,S}:$ 前$i?1$個站都已經被有且訪問過一次，然后$[i,i+p?1]$的站的訪問狀態為$S$

（$S$從高到低第$j$位代表$i+j?1$車站的訪問狀態）

$d{p}_{i,S}\to d{p}_{i+1,S^{\prime }}:$ 必須滿足$S$的最高位為$1$且車站訪問狀態翻譯過來是一樣的

也就是說原來$S$的第$j$位，變成了$S^{\prime }$中的第$j?1$位

枚舉下一個被訪問的車站，要滿足該車站的狀態原來是$0$

最后套矩陣加速

code

#include <cstdio> #include <cstring> #define mod 30031 struct matrix {int n, m;int c[130][130];matrix() {memset( c, 0, sizeof( c ) );}matrix operator * ( matrix &t ) {matrix ans;ans.n = n, ans.m = t.m;for( int i = 1;i <= n;i ++ )for( int j = 1;j <= t.m;j ++ )for( int k = 1;k <= m;k ++ )ans.c[i][j] = ( ans.c[i][j] + c[i][k] * t.c[k][j] ) % mod;return ans;} }g, ret; int n, k, p, tot; int s[130];matrix qkpow( matrix x, int y ) {matrix ans;ans.n = ans.m = x.n;for( int i = 1;i <= ans.n;i ++ )ans.c[i][i] = 1;while( y ) {if( y & 1 ) ans = ans * x;x = x * x;y >>= 1;}return ans; }int main() {scanf( "%d %d %d", &n, &k, &p );for( int i = ( 1 << p - 1 );i < ( 1 << p );i ++ ) {int cnt = 0;for( int j = 0;j < p;j ++ )if( 1 << j & i ) cnt ++;if( cnt == k ) s[++ tot] = i;}g.n = g.m = ret.m = tot, ret.n = 1;ret.c[1][tot] = 1;for( int i = 1;i <= tot;i ++ )for( int j = 1;j <= tot;j ++ ) {int New = s[i] - ( 1 << p - 1 ) << 1;for( int w = 0;w < p;w ++ )if( ! ( 1 << w & New ) && New + ( 1 << w ) == s[j] ) {g.c[i][j] = 1;break;}}g = qkpow( g, n - k );ret = ret * g;printf( "%d\n", ret.c[1][tot] );return 0; }

總結

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