SPOJ SUBLEX Lexicographical Substring Search - 后缀数组

题目传送门

　　传送门I

　　传送门II

题目大意

　　给定一个字符串，多次询问它的第$k$大本质不同的子串，输出它。

　　考虑后缀Trie。依次考虑每个后缀新增的本质不同的子串个数，显然，它是$n - sa[i] - height[i]$。

　　求出$height$数组后，求一求本质不同的子串个数的前缀和，可以对每个询问二分。

　　这里可以直接离线，$O(n + m)$扫一扫就好了。

Code

 /**
  * SPOJ
  * Problem#SUBLEX
  * Accepted
  * Time: 30ms
  * Memory: 19456k
  */
 #include <bits/stdc++.h>
 using namespace std;
 typedef bool boolean;

 typedef class Pair3 {
     public:
         int x, y, id;
 }Pair3;

 typedef class Query {
     public:
         int k, s, t, id;

         boolean operator < (Query b) const {
             return k < b.k;
         }
 }Query;

 const int N = 1e5 + , M = ;

 int n, m;
 char str[N], bs[N];
 int rk[N << ], sa[N], hei[N], cnt[N];
 Pair3 ps[N], buf[N];
 Query qs[M];

 inline void init() {
     scanf("%s", str + );
     n = strlen(str + );
     scanf("%d", &m);
     for (int i = ; i <= m; i++)
         scanf("%d", &qs[i].k), qs[i].id = i;
 }

 inline void radix_sort(Pair3* x, Pair3* y) {
     int m = ((n > ) ? (n) : ());
     memset(cnt, , sizeof(int) * (m + ));
     for (int i = ; i <= n; i++)    cnt[x[i].y]++;
     for (int i = ; i <= m; i++)    cnt[i] += cnt[i - ];
     for (int i = ; i <= n; i++)    y[cnt[x[i].y]--] = x[i];
     memset(cnt, , sizeof(int) *  (m + ));
     for (int i = ; i <= n; i++)    cnt[y[i].x]++;
     for (int i = ; i <= m; i++)    cnt[i] += cnt[i - ];
     for (int i = n; i; i--)    x[cnt[y[i].x]--] = y[i];
 }

 inline void build_sa() {
     for (int i = ; i <= n; i++)
         rk[i] = str[i];
     for (int k = , dif = ; k <= n; k <<= , dif = ) {
         for (int i = ; i <= n; i++)
             ps[i].x = rk[i], ps[i].y = rk[i + k], ps[i].id = i;
         radix_sort(ps, buf);
         rk[ps[].id] = ++dif;
         for (int i = ; i <= n; i++)
             rk[ps[i].id] = (dif += (ps[i].x != ps[i - ].x || ps[i].y != ps[i - ].y));
         if (dif == n)
             break;
     }
     for (int i = ; i <= n; i++)
         sa[rk[i]] = i;
 }

 inline void get_height() {
     for (int i = , j, k = ; i <= n; i++) {
         if (k)    k--;
         if (rk[i] > ) {
             for (j = sa[rk[i] - ]; i + k <= n && j + k <= n && str[i + k] == str[j + k]; k++);
             hei[rk[i]] = k;
         } else
             hei[] = ;
     }
 }

 inline void solve() {
     build_sa();
     get_height();
 //    for (int i = 1; i <= n; i++)
 //        cerr << hei[i] << " ";
 //    cerr << endl;
     sort(qs + , qs + m + );
     long long cur = ;
     for (int i = , nq = , delta; i <= n && nq <= m; i++) {
         delta = n - sa[i] +  - hei[i];
         if (cur + delta < qs[nq].k)
             cur += delta;
         else {
             qs[nq].s = sa[i], qs[nq].t = sa[i] + hei[i] + (qs[nq].k - cur);
             nq++, i--;
         }
     }
     for (int i = ; i <= m; i++)
         while (qs[i].id != i)
             swap(qs[qs[i].id], qs[i]);
     for (int i = ; i <= m; i++) {
         int len = qs[i].t - qs[i].s;
         memcpy(bs, str + qs[i].s , len);
         bs[len] = ;
         puts(bs);
     }
 }

 int main() {
     init();
     solve();
     return ;
 }