POJ3581---Sequence 后缀树组

题意：n个数字组成的序列，第一个数字最大，，把序列分成3部分，每个部分分别翻转，输出翻转后字典序最小的序列。。

后缀数组变一下，，先求出 第一个分割的位置，，然后再求一次后缀数组，，求出第二个位置。。输出就好了。

此题要采用单组输入。。。

 #include <set>
 #include <map>
 #include <cmath>
 #include <ctime>
 #include <queue>
 #include <stack>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 typedef unsigned long long ull;
 typedef long long ll;
 const int inf = 0x3f3f3f3f;
 const double eps = 1e-;
 const int maxn = 2e5+;
 int s[maxn], rev_s[maxn << ];
 int sa[maxn], rank[maxn], tmp[maxn];
 int k, sort_len;
 bool cmp(int i, int j)
 {
     if (rank[i] != rank[j])
         return rank[i] < rank[j];
     else
     {
         int x = i + k <= sort_len ? rank[i+k] : -;
         int y = j + k <= sort_len ? rank[j+k] : -;
         return x < y;
     }
 }
 void build_sa(int str[], int len)
 {
     sort_len = len;
     for (int i = ; i <= len; i++)
     {
         sa[i] = i;
         rank[i] = i < len ? str[i] : -;
     }
     for (k = ; k <= len; k *= )
     {
         sort(sa, sa + len + , cmp);
         tmp[sa[]] = ;
         for (int i = ; i <= len; i++)
         {
             tmp[sa[i]] = tmp[sa[i-]] + (cmp(sa[i-],sa[i]) ?  : );
         }
         for (int i = ; i <= len; i++)
             rank[i] = tmp[i];
     }
 }
 int main(void)
 {
     #ifndef ONLINE_JUDGE
         freopen("in.txt","r",stdin);
     #endif
     int n;
     //while (~scanf ("%d", &n))
     scanf ("%d", &n);
     {
         for (int i = ; i < n; i++)
             scanf ("%d", s + i);
         reverse_copy(s, s + n, rev_s);
         build_sa(rev_s, n);
         int pos1;
         for (int i = ; i <= n; i++)
         {
             pos1 = n - sa[i] - ;
             if (pos1 >=  && pos1 <= n - )
                 break;
         }
         int len = n - pos1 - ;
         reverse_copy(s+pos1+, s+n, rev_s);
         reverse_copy(s+pos1+, s+n, rev_s+len);
         build_sa(rev_s, len << );
         int pos2;
         for (int i = ; i <=  * len; i++)
         {
             pos2 = len - sa[i] - ;
             if (sa[i] < len  && pos1++pos2 < n-)
                 break;
         }
         for (int i = pos1; i >= ; i--)
             printf("%d\n",s[i]);
         for (int i = pos2+pos1+; i > pos1; i--)
             printf("%d\n",s[i]);
         for (int i = n-; i > pos2+pos1+; i--)
             printf("%d\n", s[i]);
     }
     return ;
 }