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  • [POJ3693][SP687]Maximum repetition substring/REPEATS

    参考博客1

    参考博客2,这里面输出方案的部分复杂度我不会证,而且觉得很假,我肯定95%写这种的人都是抄的而且不会证复杂度。。。。

    输出方案的代码参考黄学长的

    还有就是论文,这里面有下载链接

    记得特判长度是1的情况

    关于输出方案

    #include <iostream>
    #include <cassert>
    #include <algorithm>
    #include <cstdio>
    #include <cstdlib>
    #include <vector>
    #include <cstring>
    using namespace std;
    const int N = 1e5 + 7;
    typedef pair<int, int> pii;
    vector<pii>qq;
    int T, n, st[N][17];
    
    struct SA {
      int k, sa[N], rnk[N], H[N], st[N][17];
      char s[N];
      bool cmp(int *y, int a, int b, int m) {return y[a] == y[b] && y[a + m] == y[b + m];}
      void Sort(int *x, int *y, int *rk) {
        static int C[N];
        for (int i = 0; i <= k; ++i) C[i] = 0;
        for (int i = 1; i <= n; ++i) ++C[rk[i]];
        for (int i = 1; i <= k; ++i) C[i] += C[i - 1];
        for (int i = n; i; --i) y[C[rk[x[i]]]--] = x[i];
      }
      void get_SA() {
        static int Y[N];
        int *y = Y, *rk = rnk;
        k = 128;
        for (int i = 1; i <= n; ++i) rk[i] = s[y[i] = i];
        Sort(y, sa, rk);
        for (int m = 1, p = 0; p < n; k = p, m <<= 1) {
          for (p = 0; p < m; ++p) y[p + 1] = n - m + p + 1;
          for (int i = 1; i <= n; ++i) if (sa[i] > m) y[++p] = sa[i] - m;
          Sort(y, sa, rk), swap(rk, y);
          rk[sa[p = 1]] = 1;
          for (int i = 2; i <= n; ++i) rk[sa[i]] = cmp(y, sa[i], sa[i - 1], m) ? p : ++p;
        }
        for (int i = 1; i <= n; ++i) rnk[sa[i]] = i, Y[i] = 0;
      }
      void get_H() {
        for (int i = 1, k = 0; i <= n; H[rnk[i++]] = k)
          for (k ? --k : 0; s[i + k] == s[sa[rnk[i] - 1] + k]; ++k);
        for (int i = 2; i <= n; ++i) st[i][0] = H[i];
        for (int j = 1; j <= __lg(n); ++j)
          for (int i = 2; i + (1 << j) - 1 <= n; ++i)
            st[i][j] = min(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);
      }
      void clear() {
        fill(rnk + 1, rnk + 1 + n, 0);
      }
      int lcp(int x, int y) {
        if (x == y) return n - x + 1;
        x = rnk[x], y = rnk[y];
        if (x > y) swap(x, y);
        ++x;
        int k = __lg(y - x + 1);
        return min(st[x][k], st[y - (1 << k) + 1][k]);
      }
    } a, b;
    inline bool chmax(int &x, int y) {return x < y ? x = y, 1 : 0;}
    inline bool chmin(int &x, int y) {return x > y ? x = y, 1 : 0;}
    inline int Min(int l, int r) {
      int k = __lg(r - l + 1);
      return min(st[l][k], st[r - (1 << k) + 1][k]);
    }
    int main() {
      while (~scanf("%s", a.s + 1) && a.s[1] != '#') {
        int ans = 1, ansl, ansrk = 1e9, Len;
        n = strlen(a.s + 1);
        for (int i = 1; i <= n; ++i) b.s[i] = a.s[n - i + 1];
        b.s[n + 1] = a.s[n + 1] = 0;
        a.get_SA(), a.get_H(), b.get_SA(), b.get_H();
        for (int i = 1; i <= n; ++i) st[i][0] = a.rnk[i];
        for (int j = 1; j <= __lg(n); ++j)
          for (int i = 1; i + (1 << j) - 1 <= n; ++i)
            st[i][j] = min(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);
        for (int len = 1; len <= n; ++len)
          for (int i = 1; i + len <= n; i += len) {
            int LCP = a.lcp(i, i + len), LCS = b.lcp(n - i + 1, n - i - len + 1), tmp;
            if (chmax(ans, tmp = (LCP + LCS - 1 + len) / len)) ansrk = 1e9;
            if (tmp == ans) {
              int l = i - LCS + 1, res = (LCP + LCS - 1 + len) % len;
              if (chmin(ansrk, Min(l, l + res))) ansl = a.sa[ansrk], Len = len;
            }
          }
        static int cnt = 0; printf("Case %d: ", ++cnt);
        if (ans == 1) putchar(*min_element(a.s + 1, a.s + 1 + n));
        else for (int i = 0; i < ans * Len; ++i) putchar(a.s[ansl + i]);
        puts("");
        a.clear(), b.clear();
      }
      return 0;
    }
    
    #include <bits/stdc++.h>
    using namespace std;
    const int N = 5e4 + 7;
    
    int T, n;
    
    struct SA {
      int k, sa[N], rnk[N], H[N], st[N][17];
      char s[N];
      bool cmp(int *y, int a, int b, int m) {return y[a] == y[b] && y[a + m] == y[b + m];}
      void Sort(int *x, int *y, int *rk) {
        static int C[N];
        for (int i = 0; i <= k; ++i) C[i] = 0;
        for (int i = 1; i <= n; ++i) ++C[rk[i]];
        for (int i = 1; i <= k; ++i) C[i] += C[i - 1];
        for (int i = n; i; --i) y[C[rk[x[i]]]--] = x[i];
      }
      void get_SA() {
        static int Y[N];
        int *y = Y, *rk = rnk;
        k = 128;
        for (int i = 1; i <= n; ++i) rk[i] = s[y[i] = i];
        Sort(y, sa, rk);
        for (int m = 1, p = 0; p < n; k = p, m <<= 1) {
          for (p = 0; p < m; ++p) y[p + 1] = n - m + p + 1;
          for (int i = 1; i <= n; ++i) if (sa[i] > m) y[++p] = sa[i] - m;
          Sort(y, sa, rk), swap(rk, y);
          rk[sa[p = 1]] = 1;
          for (int i = 2; i <= n; ++i) rk[sa[i]] = cmp(y, sa[i], sa[i - 1], m) ? p : ++p;
        }
        for (int i = 1; i <= n; ++i) rnk[sa[i]] = i;
      }
      void get_H() {
        for (int i = 1, k = 0; i <= n; H[rnk[i++]] = k)
          for (k ? --k : 0; s[i + k] == s[sa[rnk[i] - 1] + k]; ++k);
        for (int i = 2; i <= n; ++i) st[i][0] = H[i];
        for (int j = 1; j <= __lg(n); ++j)
          for (int i = 2; i + (1 << j) - 1 <= n; ++i) st[i][j] = min(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);
      }
      int lcp(int x, int y) {
        x = rnk[x], y = rnk[y];
        if (x > y) swap(x, y);
        ++x;
        int k = __lg(y - x + 1);
        return min(st[x][k], st[y - (1 << k) + 1][k]);
      }
    } a, b;
    inline int gi() {int c = getchar(); while (!isalpha(c)) c = getchar(); return c;}
    inline void chmax(int &x, int y) {if (x < y) x = y;}
    int main() {
      cin >> T;
      while (T--) {
        cin >> n;
        for (int i = 1; i <= n; ++i) a.s[i] = gi();
        for (int i = a.s[n + 1] = 0; i <= n; ++i) b.s[i] = a.s[n - i + 1];
        a.get_SA(), a.get_H(), b.get_SA(), b.get_H();
        int ans = 1;
        for (int len = 1; len <= n; ++len)
          for (int i = 1; i + len <= n; i += len) chmax(ans, (a.lcp(i, i + len) + b.lcp(n - i + 1, n - i - len + 1) + len - 1) / len);
        cout << ans << endl;
      }
      return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/storz/p/10620075.html
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