Codeforces 696 D. Legen ...

Source: Internet
Author: User


Each string has some value, asking how much value it can generate for a string of length \ (l\), the total number of characters not exceeding \ (200\), \ (L\leqslant 10^{14}\).


AC automata + multiply Floyd.

Use the AC automaton to count the weights that are obtained at each node.

Then in the AC automata from the root node to find a longest road, the Floyd multiplied on it ...

Find oneself AC automatic machine do fail tree when write wrong, changed for a long time did not change out, finally found can't directly throw the root into the queue ... All child nodes of the root should be thrown in.

Then you can do it like a quick power. Write the matrix when you need to determine whether to arrive, I use \ ( -1\) to indicate whether to arrive.

Transfer is \ (c[i][j]=max\{c[i][j],a[i][k]+b[k][j]\},a[i][k]!=-1,b[k][j]!=-1\).

#include <bits/stdc++.h>using namespace Std;const int M = 205;const int N = 20050;typedef long Long ll;typedef vecto r< LL > Vec;typedef vector< Vec > Mat;void F (Mat &a,int v=-1) {int n=a.size (); for (int. i=0;i<n;i++) for ( int j=0;j<n;j++) A[i][j]=v;} MAT operator * (const MAT &a,const mat &b) {int n=a.size (); Mat C (N,vec (n)); F (C); for (int k=0;k<n;k++) for (int. i=0;i<n;i++) for (int j=0;j<n;j++) if (a[i][k]!=-1 && b[k][j]!=-1) c[i ][j]=max (C[i][j],a[i][k]+b[k][j]); return C;} Mat operator ^ (Mat a,ll b) {int n=a.size (); Mat R (N,vec (n)); F (R); for (int i=0;i<n;i++) r[i][i]=0;for (; b;b>>=1,a=a*a) if (b&1) R=r*a;return R;} void Print (const Mat &a) {int n=a.size (); for (int. i=0;i<n;i++) {for (int j=0;j<n;j++) cout<<a[i][j]< < ""; Cout<<endl;} cout<< "---------------------" &LT;&LT;ENDL;} struct Ac_auto {int rt,cnt;int f[n],ch[n][26],v[n];int NewNode () {++cnt;v[cnt]=f[cnt]=0,memset (ch[cnt],0,sizeof (ch[ CNT]); return CNT; }void init () {Cnt=rt=f[rt]=v[rt]=0;memset (ch[rt],0,sizeof (Ch[rt]));} void Insert (char *s,int val) {int L=strlen (s), o=rt;for (int i=0;i<l;i++) {if (!ch[o][s[i]-' a ']) ch[o][s[i]-' a ']= NewNode (); o=ch[o][s[i]-' A '];} V[o]+=val;} void Getfail () {queue< int > q;for (int i=0;i<26;i++) if (Ch[rt][i]) Q.push (Ch[rt][i]); for (;! Q.empty ();) {int X=q.front (); Q.pop (); for (int i=0;i<26;i++) {if (!ch[x][i]) Ch[x][i]=ch[f[x]][i];else Q.push (Ch[x][i]), f[ch[x] [I]] =ch[f[x]][i],v[ch[x][i]]+=v[f[ch[x][i]];}} for (int i=0;i<=cnt;i++) cout<<i<< "--" &LT;&LT;F[I]&LT;&LT;ENDL;} void make (Mat &a) {a.resize (Cnt+1,vec (cnt+1)); int n=a.size (); F (A); for (int. i=0;i<n;i++) for (int j=0;j<26;j++) a[i][ch[i][j]]=v[ch[i][j]],a[i][ch[rt][j]]=v[ch[rt][j];}} Ac LL N,l,ans; LL A[n];char S[n];int Main () {//freopen ("", "R", stdin); Ios::sync_with_stdio (false); Ac.init ();cin>>n> >l;for (int i=1;i<=n;i++) cin>>a[i];for (int i=1;i<=n;i++) Cin>>s,ac.insert (S,a[i]); Ac.getFail (); Mat R;ac. Make (R);//print (R); r=r^l;for (int i=0;i< (int) r.size (); i++) Ans=max (Ans,r[0][i]);cout<<ans<<endl; return 0;}


Codeforces 696 D. Legen ...

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