题目传送门

sol:先通过AC自动机构建字典，用$dp[i]$表示长串前$i$位的最小代价，若有一个单词$s$是长串的前$i$项的后缀，那么可以用$dp[i - len(s)] + val(s)$转移到$dp[i]$。

• AC自动机

  #include <bits/stdc++.h>
  using namespace std;
  typedef long long LL;
  typedef pair<int, int> PII;
  const int MAXN = 500010;
  const LL INF = 0x3f3f3f3f3f3f3f3f;
  struct Trie {
      int son[MAXN][26], fail[MAXN], last[MAXN], dep[MAXN];
      LL val[MAXN], dp[MAXN]; int tot, root;
      int add_node(int d) {
          memset(son[tot], -1, sizeof(son[tot]));
          dep[tot] = d; val[tot] = INF;
          return tot ++;
      }
      void init() {
          memset(dp, INF, sizeof(dp));
          dp[0] = tot = 0;
          root = add_node(0);
      }
      void insert(char* s, LL v) {
          int p = root;
          for (int i = 0; s[i]; i++) {
              int index = s[i] - 'a';
              if (son[p][index] == -1)
                  son[p][index] = add_node(i + 1);
              p = son[p][index];
          }
          val[p] = min(val[p], v);
      }
      void build() {
          queue<int> que; fail[root] = root;
          for (int i = 0; i < 26; i++) {
              if (son[root][i] == -1) son[root][i] = root;
              else {
                  fail[son[root][i]] = root;
                  last[son[root][i]] = root;
                  que.push(son[root][i]);
              }
          }
          while (!que.empty()) {
              int p = que.front(); que.pop();
              for (int i = 0; i < 26; i++) {
                  if (son[p][i] == -1) son[p][i] = son[fail[p]][i];
                  else {
                      fail[son[p][i]] = son[fail[p]][i];
                      if (val[son[fail[p]][i]] != INF)
                          last[son[p][i]] = son[fail[p]][i];
                      else
                          last[son[p][i]] = son[last[p]][i];
                      que.push(son[p][i]);
                  }
              }
          }
      }
      LL slove(char* s) {
          int p = root, len = strlen(s);
          for (int i = 0; i < len; i++) {
              int index = s[i] - 'a';
              p = son[p][index];
              for (int tmp = p; tmp != root; tmp = last[tmp])
                  dp[i + 1] = min(dp[i + 1], dp[i + 1 - dep[tmp]] + val[tmp]);
          }
          if (dp[len] == INF) return -1;
          else return dp[len];
      }
  } ac;
  char s[MAXN];
  int main() {
      int n, v;
      scanf("%d", &n); ac.init();
      for (int i = 1; i <= n; i++) {
          scanf("%s%d", s, &v);
          ac.insert(s, v);
      }
      ac.build();
      scanf("%s", s);
      printf("%lld\n", ac.slove(s));
      return 0;
  }

  AC自动机不熟练，硬怼了三天才敲出来，一直超时。又学到了两个AC自动机的优化。