[JSOI2008]火星人

嘟嘟嘟




嗯。




splay维护哈希。

如题，用splay维护哈希，查找的时候二分。所以复杂度是取决于询问复杂度：\(O(n \log^ 2{n})\)。

这道题还有一个技巧，就是一个节点记录的是他的子树的哈希值，所以树的的形态改变的同时，每一个节点记录的哈希值也在改变。在pushup的时候，应该这么写：\(t[now].has = t[l].has * base ^ {t[r].siz + 1} + (s[now] - 'a') * base ^ {t[r].siz} + t[r].has\)。




然后好像就没什么好说的啦，各种操作的中心思想都是通过旋转前驱后继提取区间。

然后注意\(base\)别取字符集大小，我因为取了\(26\)导致大的点没过去，应该是取的太小导致冲突的概率比较大吧。

然而取\(27\)就过了～

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int m;
ull p[maxn];
char s[maxn], c[2], tp[2];

struct Tree
{
  int ch[2], fa;
  int siz; ull has;
}t[maxn];
int root, ncnt = 0;

void _PrintTr(int now)
{
  if(!now) return;
  printf("nd:%d ls:%d rs:%d\n", now, t[now].ch[0], t[now].ch[1]);
  _PrintTr(t[now].ch[0]); _PrintTr(t[now].ch[1]);
}

void pushup(int now)
{
  int l = t[now].ch[0], r = t[now].ch[1];
  t[now].siz = t[l].siz + t[r].siz + 1;
  t[now].has = t[l].has * p[t[r].siz + 1] + (s[now] - 'a') * p[t[r].siz] + t[r].has;
}
void rotate(int x)
{
  int y = t[x].fa, z = t[y].fa, k = (t[y].ch[1] == x);
  t[z].ch[t[z].ch[1] == y] = x; t[x].fa = z;
  t[y].ch[k] = t[x].ch[k ^ 1]; t[t[y].ch[k]].fa = y;
  t[x].ch[k ^ 1] = y; t[y].fa = x;
  pushup(y); pushup(x);
}
void splay(int x, int s)
{
  while(t[x].fa != s)
    {
      int y = t[x].fa, z = t[y].fa;
      if(z != s)
	{
	  if((t[z].ch[0] == y) ^ (t[y].ch[0] == x)) rotate(x);
	  else rotate(y);
	}
      rotate(x);
    }
  if(!s) root = x;
}
void build(int L, int R, int f)
{
  if(L > R) return;
  int mid = (L + R) >> 1;
  if(L == R) t[L].siz = 1, t[L].has = s[L] - 'a';
  else build(L, mid - 1, mid), build(mid + 1, R, mid);
  t[f].ch[mid > f] = mid; t[mid].fa = f;
  pushup(mid);
}
int rnk(int k)
{
  int now = root;
  while(1)
    {
      if(k <= t[t[now].ch[0]].siz) now = t[now].ch[0];
      else if(k == t[t[now].ch[0]].siz + 1) return now;
      else k -= (t[t[now].ch[0]].siz + 1), now = t[now].ch[1];
    }
}
void update(int x)
{
  int now = rnk(x); splay(now, 0);
  s[now] = tp[0];
  pushup(now);
}
void insert(int x)
{
  int a = rnk(x), b = rnk(x + 1);
  splay(a, 0); splay(b, a);
  s[++ncnt] = tp[0];
  t[ncnt].has = tp[0] - 'a';
  t[ncnt].siz = 1;
  t[ncnt].ch[0] = t[ncnt].ch[1] = 0;
  t[b].ch[0] = ncnt; t[ncnt].fa = b;
  pushup(b); pushup(a);
}
bool judge(int x, int y, int len)
{
  if(x + len > ncnt || y + len > ncnt) return 0;
  int a = rnk(x - 1), b = rnk(x + len);
  splay(a, 0); splay(b, a);
  //_PrintTr(root);
  //puts("~~~~~~~~~~~~~~~~~~");
  ull has1 = t[t[b].ch[0]].has;
  a = rnk(y - 1), b = rnk(y + len);
  splay(a, 0); splay(b, a);
  //_PrintTr(root);
  ull has2 = t[t[b].ch[0]].has;
  return has1 == has2;
}
int query(int x, int y)
{
  int L = 0, R = ncnt;
  while(L < R)
    {
      int mid = (L + R + 1) >> 1;
      if(judge(x, y, mid)) L = mid;
      else R = mid - 1;
    }
  return L;
}

int main()
{
  p[0] = 1; for(int i = 1; i < maxn; ++i) p[i] = p[i - 1] * 47; //学姐强啊
  scanf("%s", s + 2); ncnt = strlen(s + 2) + 2;
  s[1] = s[ncnt] = 'f';
  build(1, ncnt, 0);
  for(int i = 1; i <= ncnt && !root; ++i) if(!t[i].fa) root = i;
  m = read();
  for(int i = 1, x; i <= m; ++i)
    {
      scanf("%s", c); x = read() + 1;
      if(c[0] == 'Q')
	{
	  int y = read() + 1;
	  write(query(x, y)), enter;
	}
      else
	{
	  scanf("%s", tp);
	  if(c[0] == 'R') update(x);
	  else insert(x);
	}
      //_PrintTr(root);
    }
  return 0;
}