Longest Common Substring II SPOJ

题意：多个串的最大公共子串
思路：用广义后缀数组构造后，记录每个点包含的节点，最后判断那个点包含所有点。

#include <bits/stdc++.h>
using namespace std;
#define MAXN 2000000  // 双倍字符串长度
#define CHAR_NUM 30   // 字符集个数，注意修改下方的 (-'a')
#define NUM 15        // 字符串个数
struct exSAM {
  int len[MAXN];             // 节点长度
  int link[MAXN];            // 后缀链接，link
  int next[MAXN][CHAR_NUM];  // 转移
  int tot;                   // 节点总数：[0, tot)
  int lenSorted[MAXN];   // 按照 len 排序后的数组，仅排序 [1, tot)
                         // 部分，最终下标范围 [0, tot - 1)
  int sizeC[MAXN][NUM];  // 表示某个字符串的子串个数
  int curString;         // 字符串实际个数
  /**
   * 计数排序使用的辅助空间数组
   */
  int lc[MAXN];  // 统计个数
  void init() {
    tot = 1;
    link[0] = -1;
  }
  int insertSAM(int last, int c) {
    int cur = next[last][c];
    len[cur] = len[last] + 1;
    int p = link[last];
    while (p != -1) {
      if (!next[p][c])
        next[p][c] = cur;
      else
        break;
      p = link[p];
    }
    if (p == -1) {
      link[cur] = 0;
      return cur;
    }
    int q = next[p][c];
    if (len[p] + 1 == len[q]) {
      link[cur] = q;
      return cur;
    }
    int clone = tot++;
    for (int i = 0; i < CHAR_NUM; ++i)
      next[clone][i] = len[next[q][i]] != 0 ? next[q][i] : 0;
    len[clone] = len[p] + 1;
    while (p != -1 && next[p][c] == q) {
      next[p][c] = clone;
      p = link[p];
    }
    link[clone] = link[q];
    link[cur] = clone;
    link[q] = clone;
    return cur;
  }
  int insertTrie(int cur, int c) {
    if (!next[cur][c]) next[cur][c] = tot++;
    sizeC[next[cur][c]][curString]++;
    return next[cur][c];
  }
  void insert(const string &s) {
    int root = 0;
    for (auto ch : s) root = insertTrie(root, ch - 'a');
    curString++;
  }
  void insert(const char *s, int n) {
    int root = 0;
    for (int i = 0; i < n; ++i) root = insertTrie(root, s[i] - 'a');
    curString++;
  }
  void build() {
    queue<pair<int, int>> q;
    for (int i = 0; i < 26; ++i)
      if (next[0][i]) q.push({i, 0});
    while (!q.empty()) {
      auto item = q.front();
      q.pop();
      auto last = insertSAM(item.second, item.first);
      for (int i = 0; i < 26; ++i)
        if (next[last][i]) q.push({i, last});
    }
  }
  void sortLen() {
    for (int i = 1; i < tot; ++i) lc[i] = 0;
    for (int i = 1; i < tot; ++i) lc[len[i]]++;
    for (int i = 2; i < tot; ++i) lc[i] += lc[i - 1];
    for (int i = 1; i < tot; ++i) lenSorted[--lc[len[i]]] = i;
  }
  void getSizeLen() {
    for (int i = tot - 2; i >= 0; --i)
      for (int j = 0; j < curString; ++j)
        sizeC[link[lenSorted[i]]][j] += sizeC[lenSorted[i]][j];
  }
  void debug() {
    cout << "     i      len      link       ";
    for (int i = 0; i < 26; ++i) cout << "  " << (char)('a' + i);
    cout << endl;
    for (int i = 0; i < tot; ++i) {
      cout << "i: " << setw(3) << i << " len: " << setw(3) << len[i]
           << " link: " << setw(3) << link[i] << " Next: ";
      for (int j = 0; j < CHAR_NUM; ++j) {
        cout << setw(3) << next[i][j];
      }
      cout << endl;
    }
  }
} exSam;
int main() {
  exSam.init();
  string s;
  while (cin >> s) exSam.insert(s);
  exSam.build();
  exSam.sortLen();
  exSam.getSizeLen();
  int ans = 0;
  for (int i = 0; i < exSam.tot; ++i) {
    bool flag = true;
    for (int j = 0; j < exSam.curString; ++j) {
      if (!exSam.sizeC[i][j]) {
        flag = false;
        break;
      }
    }
    if (flag) ans = max(ans, exSam.len[i]);
  }
  cout << ans << endl;
}