我们能很容易写出在AC自动机上的dp, dp[ k ][ i ][ j ]表示走了 k 步从 i 走到 j 的最大值。
k 很大我们考虑矩阵优化, 直接搞就好啦。
不知道为什么在本机上M * M * M * M **** * M 打多了就会卡死不知道为什么。。。
#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 200 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 998244353; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;} int n, a[N]; LL l; char s[N][N]; int MN; struct Matrix { LL a[207][207]; Matrix() { for(int i = 0; i < MN; i++) for(int j = 0; j < MN; j++) a[i][j] = 0; } void init() { for(int i = 0; i < MN; i++) for(int j = 0; j < MN; j++) a[i][j] = -INF; } Matrix operator * (const Matrix &B) const { Matrix C; C.init(); for(int i = 0; i < MN; i++) for(int j = 0; j < MN; j++) for(int k = 0; k < MN; k++) chkmax(C.a[i][j], a[i][k] + B.a[k][j]); return C; } Matrix operator ^ (LL b) { Matrix C = (*this); Matrix A = (*this); b--; while(b) { if(b & 1) C = C * A; A = A * A; b >>= 1; } return C; } } M; struct Ac { // 1. init before use 2. cheke character set int ch[N][26], f[N], tot, sz, subVal; int val[N]; inline int newNode() { tot++; f[tot] = 0; memset(ch[tot], 0, sizeof(ch[tot])); return tot; } void init(int _sz, int _subVal) { sz = _sz; subVal = _subVal; tot = -1; newNode(); } inline int idx(int c) {return c - subVal;} void addStr(char* s, int w) { int u = 0; for(int i = 0; s[i]; i++) { int c = idx(s[i]); if(!ch[u][c]) ch[u][c] = newNode(); u = ch[u][c]; } val[u] += w; } void build() { queue<int> que; for(int c = 0; c < sz; c++) { int v = ch[0][c]; if(!v) ch[0][c] = 0; else f[v] = 0, que.push(v); } while(!que.empty()) { int u = que.front(); que.pop(); val[u] += val[f[u]]; for(int c = 0; c < sz; c++) { int v = ch[u][c]; if(!v) ch[u][c] = ch[f[u]][c]; else f[v] = ch[f[u]][c], que.push(v); } } } void getMatrix() { MN = tot + 1; for(int i = 0; i < MN; i++) for(int j = 0; j < MN; j++) M.a[i][j] = -INF; for(int u = 0; u <= tot; u++) { for(int c = 0; c < sz; c++) { int v = ch[u][c]; if(!v) continue; chkmax(M.a[u][v], val[v]); } } } } ac; int main() { ac.init(26, 'a'); scanf("%d%lld", &n, &l); for(int i = 1; i <= n; i++) scanf("%d", &a[i]); for(int i = 1; i <= n; i++) scanf("%s", s[i]); for(int i = 1; i <= n; i++) ac.addStr(s[i], a[i]); ac.build(); ac.getMatrix(); M = M ^ l; LL ans = 0; for(int i = 0; i < MN; i++) chkmax(ans, M.a[0][i]); printf("%lld ", ans); return 0; } /* */