poj 3294 Life Forms - 后缀数组 - 二分答案

题目传送门

　　传送门I

　　传送门II

题目大意

　　给定$n$个串，询问所有出现在严格大于$\frac{n}{2}$个串的最长串。不存在输出'?'

　　用奇怪的字符把它们连接起来。然后求sa，hei，二分答案，按mid分组。

　　判断每一组存在的后缀属于的原串的种类数是不是存在那么多个。

　　这个做法可以推广到多串求LCS，然后多个log，完美在SPOJ上T掉。

Code

 /**
  * poj
  * Problem#3294
  * Accepted
  * Time: 391ms
  * Memory: 5024k
  */
 #include <iostream>
 #include <cstring>
 #include <cstdlib>
 #include <cstdio>
 #ifndef WIN32
 #define Auto "%lld"
 #else
 #define Auto "%I64d"
 #endif
 using namespace std;
 typedef bool boolean;
 #define ll long long

 #define pii pair<int, int>
 #define fi first
 #define sc second

 const int N = 1e5 + , M = ;

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

         Pair3() {    }
         Pair3(int x, int y, int id):x(x), y(y), id(id) {    }
 }Pair3;

 typedef class SuffixArray {
     protected:
         Pair3 T1[N], T2[N];
         int cnt[N];

     public:
         int n;
         char *str;
         int sa[N], rk[N], hei[N];

         void set(int n, char* str) {
             this->n = n;
             this->str = str;
             memset(sa, , sizeof(sa));
             memset(rk, , sizeof(rk));
             memset(hei, , sizeof(hei));
         }

         void radix_sort(Pair3* x, Pair3* y) {
             int m = max(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];
         }

         void build() {
             for (int i = ; i < n; i++)
                 rk[i] = str[i];
             for (int k = ; k < n; k <<= ) {
                 for (int i = ; i + k < n; i++)
                     T1[i] = Pair3(rk[i], rk[i + k], i);
                 for (int i = n - k; i < n; i++)
                     T1[i] = Pair3(rk[i], , i);
                 radix_sort(T1, T2);
                 int diff = ;
                 rk[T1[].id] = ;
                 for (int i = ; i < n; i++)
                     rk[T1[i].id] = (T1[i].x == T1[i - ].x && T1[i].y == T1[i - ].y) ? (diff) : (++diff);
                 if (diff == n - )
                     break;
             }
             for (int i = ; i < n; i++)
                 sa[rk[i]] = i;
         }

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

         const int& operator [] (int p) {
             return sa[p];
         }

         const int& operator () (int p) {
             return hei[p];
         }
 }SuffixArray;

 int n, m, K;
 char S[N];
 int suf[N];
 SuffixArray sa;

 inline boolean init() {
     scanf("%d", &m);
     if (!m)
         return false;
     K = (m >> ) + ;
     scanf("%s", S);
     n = strlen(S);
     for (int i = ; i < m; i++) {
         S[n++] = '?';
         scanf("%s", S + n);
         n += strlen(S + n);
     }
     suf[n] = ;
     for (int i = n - ; ~i; i--)
         suf[i] = suf[i + ] + (S[i] == '?');
     sa.set(n, S);
     return true;
 }

 int chk_clock = ;
 int vis[M];

 boolean check(int L, int R, int mid) {    // [L, R)
     if (R - L < K)
         return false;
     if (suf[sa[L]] - suf[sa[L] + mid])
         return false;
     chk_clock++;
     int rt = ;
     for (int i = L; i < R && rt < K; i++)
         if (vis[suf[sa[i]]] != chk_clock)
             vis[suf[sa[i]]] = chk_clock, rt++;
     return rt >= K;
 }

 boolean check(int mid) {
     int lst = ;
     boolean rt = false;
     for (int i = ; i < n && !rt; i++)
         if (sa(i) < mid)
             rt |= check(lst, i, mid), lst = i;
     if (!rt)
         rt |= check(lst, n, mid);
     return rt;
 }

 void output(int L, int R, int res) {
     if (!check(L, R, res))
         return;
     for (int i = sa[L], j = ; j < res; j++)
         putchar(S[i + j]);
     putchar('\n');
 }

 inline void solve() {
     chk_clock = ;
     memset(vis, , sizeof(vis));

     sa.build();
     sa.get_height();

     int l = , r = n / m, mid;
     while (l <= r) {
         mid = (l + r) >> ;
         if (check(mid))
             l = mid + ;
         else
             r = mid - ;
     }

     if (!(--l))
         puts("?");
     else {
         int lst = ;
         for (int i = ; i < n; i++)
             if (sa(i) < l)
                 output(lst, i, l), lst = i;
         output(lst, n, l);
     }
     puts("");
 }

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