2009那篇论文里介绍的做法

这里面有下载链接

网上很多题解把分隔符弄成了'#'+i 这个其实可以卡掉 因为会跟正常的字母冲突

#include <algorithm>
#include <vector>
#include <cstring>
#include <cstdio>
#include <bitset>
using namespace std;

const int N = 100107;
int cnt[N], sa[N], rnk[N], H[N], n, k, s[N], bel[N], T;
char s1[1007];
bitset<107>vis;
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];
}

inline bool cmp(int *y, int a, int b, int m) {return y[a] == y[b] && y[a + m] == y[b + m];}

void get_SA() {
  static int Y[N];
  int *y = Y, *rk = rnk;
  for (int i = 1; i <= n; ++i) rk[i] = s[y[i] = i];
  k = 300;
  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(y, rk);
    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);
}
vector<int>p;
bool check(int k) {
  p.clear(), vis.reset();
  for (int i = 2; i <= n + 1; ++i) {
    if (H[i] < k) {
      if (vis.count() > T / 2) p.push_back(sa[i - 1]);
      vis.reset();
    }
    else vis[bel[sa[i]]] = vis[bel[sa[i - 1]]] = 1;
  }
  return p.size();
}

void solve() {
  n = 0;
  for (int i = 1; i <= T; ++i) {
    scanf("%s", s1 + 1);
    for (int j = 1; s1[j]; ++j) s[++n] = s1[j], bel[n] = i;
    s[++n] = 123 + i, bel[n] = 123 + i;
  }
  get_SA(), get_H();
  H[n + 1] = s[n + 1] = 0;
  int l = 0, r = 1000, mid;
  while (l <= r) {
    if (check(mid = (l + r) >> 1)) l = mid + 1;
    else r = mid - 1;
  }
  static bool f = 0;
  if (f) puts("");
  f = 1;
  if (!l) puts("?");
  else {
    check(l - 1);
    for (int i = 0; i < (int)p.size(); ++i, puts(""))
      for (int j = p[i]; j < p[i] + l - 1; ++j) putchar(s[j]);
  }
  fill(rnk + 1, rnk + 1 + n, 0);
  fill(cnt + 1, cnt + 1 + n, 0);
}

int main() {
  p.reserve(100);
  while (~scanf("%d", &T) && T) solve();
  return 0;
}