bzoj 2434 阿狸的打字机 fail树的性质

如果a串是另b串的后缀，那么在trie图上沿着b的fail指针走一定可以走到a串。

而a串在b串里出现多少次就是它是多少个前缀的后缀。

所以把fail边反向建树维护个dfs序就行了。

并不是很难。。。但没想出来TAT

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #define N 200005
 #define inf 0x3f3f3f3f
 using namespace std;
 int n;
 int ch[N][], fa[N];
 int a[N];
 int mn[N],zhi[N],size[N];
 void push_up(int x)
 {
     size[x] = size[ch[x][]] + size[ch[x][]] + ;
     mn[x] = min(zhi[x], min(mn[ch[x][]], mn[ch[x][]])); return;
 }
 int rev[N];
 void push_down(int x)
 {
     if (rev[x])
     {
         rev[x] ^= ; rev[ch[x][]] ^= ; rev[ch[x][]] ^= ;
         swap(ch[x][], ch[x][]);
     }
     return;
 }
 void rotate(int p)
 {
     int q = fa[p], y = fa[q], x = (ch[q][] == p);
     ch[q][x] = ch[p][x ^ ]; fa[ch[q][x]] = q;
     ch[p][x ^ ] = q; fa[q] = p;
     fa[p] = y;
     if (y)
     {
         if (ch[y][] == q)ch[y][] = p;
         else ch[y][] = p;
     }
     push_up(q); return;
 }
 int root;
 void splay(int x,int yy)
 {
     for (int y; y = fa[x]; rotate(x))
     {
         if (y == yy)break;
         if (fa[y] != yy)
         {
             if ((ch[fa[y]][] == y) ^ (ch[y][] == x))rotate(x);
             else rotate(y);
         }
     }
     push_up(x);
     if (!yy)root = x;
 }
 int cnt;
 int find(int k, int x)
 {
     push_down(k);
     if (mn[ch[k][]] == x)
     {
         return find(ch[k][], x);
     }
     if (zhi[k] == x)return size[ch[k][]] + ;
     return find(ch[k][], x) + size[ch[k][]] + ;
 }
 int fd(int k, int x)
 {
     push_down(k);
     if (size[ch[k][]] +  == x)return k;
     if (size[ch[k][]] +  >= x)return fd(ch[k][],x);
     return fd(ch[k][], x - size[ch[k][]] - );
 }
 struct node
 {
     int yuan,z;
     friend bool operator < (node aa,node bb)
     {
         if(aa.z!=bb.z)return aa.z<bb.z;
         return aa.yuan<bb.yuan;
     }
 }s[N];
 int yin[N];
 int main()
 {
     scanf("%d", &n); mn[] = inf;
     for (int i = ; i <= n; i++)
     {
         s[i].yuan=i;scanf("%d",&s[i].z);
     }
     sort(s+,s+n+);
     for(int i=;i<=n;i++)
     {
         a[s[i].yuan]=i;
     }
     root = ; a[n + ] = inf;
     ch[][] = ; zhi[] = a[]; size[] = n + ;
     for (int i = ; i <= n + ; i++)
     {
         size[i] = n - i + ;
         zhi[i] = a[i]; fa[i] = i - ; ch[i - ][] = i;
     }
     for (int i = n + ; i >= ; i--)push_up(i);
     splay(n, );
     for (int i = ; i <= n; i++)
     {
         int y = find(root, mn[root]);
         if(i!=n)printf("%d ",y+(i-));
         else printf("%d",y+(i-));
         int t = fd(root, y+);
         splay(t, );
         rev[ch[t][]] ^= ;
         t = fd(root, );
         splay(t, );
         ch[t][] = ;
         push_up(t);
     }
     puts("");
     return ;
 }