牛客-回文串

题目传送门

sol:根据题意，可以把字符串截成两段，要求两段中最大的两个回文串相加最大的方案，于是跑马拉车维护一个前缀，再倒着跑马拉车维护后缀。因为最优方案肯定是在两个字母之间截，所以最后只考虑在'#'位置截断的情况

• 马拉车

  #include <bits/stdc++.h>
  using namespace std;
  typedef long long LL;
  typedef pair<int, int> PII;
  const int MAXN = 2e5 + 10;
  char s[MAXN * 2];
  int p[MAXN * 2];
  int pre[MAXN * 2], suf[MAXN * 2];
  int n, k;
  void manacher() {
      k = 0; memset(p, 0, sizeof(p));
      for (int i = 1; i < n; i++) {
          if (i < p[k] + k) {
              p[i] = min(k + p[k] - i, p[2 * k - i]);
          }
          while (s[i + p[i]] == s[i - p[i]]) {
              p[i] ++;
              pre[i + p[i] - 1] = max(pre[i + p[i] - 1], p[i] - 1);
          }
          pre[i] = max(pre[i], pre[i - 1]);
      }
      k = n; memset(p, 0, sizeof(p));
      for (int i = n - 1; i > 0; i--) {
          if (i > p[k] - k) {
              p[i] = max(0, min(i - k + p[k], p[2 * k - i]));
          }
          while (s[i + p[i]] == s[i - p[i]]) {
              p[i] ++;
              suf[i - p[i] + 1] = max(suf[i - p[i] + 1], p[i] - 1);
          }
          suf[i] = max(suf[i], suf[i + 1]);
      }
  }
  int main() {
      scanf("%s", s);
      n = strlen(s);
      for (int i = n; i >= 0; i--) {
          s[i + 1 << 1] = s[i];
          s[i << 1 | 1] = '#';
      }
      s[0] = '$';
      n = n + 1 << 1;
      manacher();
  /*
      puts(s);
      for (int i = 0; i < n; i++) printf("%x", p[i]);
      cout << endl;
      for (int i = 0; i < n; i++) printf("%x", pre[i]);
      cout << endl;
      for (int i = 0; i < n; i++) printf("%x", suf[i]);
      cout << endl;
  */
      int res = 0;
      for (int i = 3; i <= n - 3; i += 2) res = max(pre[i] + suf[i], res);
      printf("%d\n", res);
      return 0;
  }