【题意】给出一个长度为n的01串，有m个操作。操作有以下几种：
1 p c在第p个字符后插入c
2 p 删除第p个字符
3 p1 p2 翻转p1到p2的字符串
4 p1 p2 求从p1开始和p2开始的两个后缀的LCP
【解题方法】
参见白书。利用字符串hash，可以维护一个区间的字符串的hash值，利用伸展树可以轻松的完成前3种操作，对于操作4，二分一下长度，将取出的两个串的hash值进行比较，就能判断是否相同。
【PS】这题有点恶心，在LCP提取区间的时候，把Rotateto写成了Rotate，查了半天错才知道。
【代码君】
//
//Created by just_sort 2016/10/15
//Copyright (c) 2016 just_sort.All Rights Reserved
//
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;
const int pp = 239641;
const int maxn = 555555;
uLL p[maxn],Hash[maxn][2];
char s[maxn];
int ch[maxn][2],pre[maxn],val[maxn],sz[maxn],rev[maxn],root,tot;
void Treaval(int x) {
if(x) {
Treaval(ch[x][0]);
printf("结点%2d:左儿子 %2d 右儿子 %2d 父结点 %2d size = %2d ,val = %2d\n",x,ch[x][0],ch[x][1],pre[x],sz[x],val[x]);
Treaval(ch[x][1]);
}
}
void debug() {printf("%d\n",root);Treaval(root);}
void Newnode(int &rt,int fa,int k)
{
rt = ++tot;
ch[rt][0] = ch[rt][1] = rev[rt] = 0;
sz[rt] = 1;
pre[rt] = fa;
Hash[rt][0] = Hash[rt][1] = val[rt] = k;
}
void pushup(int rt)
{
sz[rt] = sz[ch[rt][0]] + sz[ch[rt][1]] + 1;
int len = sz[ch[rt][0]];
Hash[rt][0] = Hash[ch[rt][0]][0] + val[rt]*p[len] + Hash[ch[rt][1]][0]*p[len+1];
len = sz[ch[rt][1]];
Hash[rt][1] = Hash[ch[rt][1]][1] + val[rt]*p[len] + Hash[ch[rt][0]][1]*p[len+1];
}
void f(int rt)
{
if(!rt) return ;
swap(ch[rt][0], ch[rt][1]);
swap(Hash[rt][0], Hash[rt][1]);
rev[rt] ^= 1;
}
void pushdown(int rt)
{
if(rev[rt]){
f(ch[rt][0]);
f(ch[rt][1]);
rev[rt] = 0;
}
}
void Rotate(int x,int kind)
{
int y = pre[x];
pushdown(y); pushdown(x);
ch[y][kind^1] = ch[x][kind];
pre[ch[x][kind]] = y;
if(pre[y]) ch[pre[y]][ch[pre[y]][1] == y] = x;
pre[x] = pre[y];
ch[x][kind] = y;
pre[y] = x;
pushup(y);
}
void Splay(int r,int goal)
{
pushdown(r);
while(pre[r]!=goal)
{
if(pre[pre[r]]==goal)
{
Rotate(r,ch[pre[r]][0]==r);
pushdown(pre[r]);
pushdown(r);
}
else
{
pushdown(pre[pre[r]]);
pushdown(pre[r]);
pushdown(r);
int y=pre[r];
int kind=(ch[pre[y]][0]==y);
if(ch[y][kind]==r)
{
Rotate(r,!kind);
Rotate(r,kind);
}
else
{
Rotate(y,kind);
Rotate(r,kind);
}
}
}
pushup(r);
if(goal==0) root=r;
}
void RotateTo(int k, int goal)
{
int rt = root;
pushdown(rt);
while(sz[ch[rt][0]] != k)
{
if(sz[ch[rt][0]] > k)
rt = ch[rt][0];
else{
k -= (sz[ch[rt][0]] + 1);
rt = ch[rt][1];
}
pushdown(rt);
}
Splay(rt,goal);
}
void Insert(int k, int c)
{
RotateTo(k, 0);
RotateTo(k+1, root);
int rt = ch[root][1];
Newnode(ch[rt][0], rt, c);
pushup(rt);
pushup(root);
}
void Delete(int p)
{
RotateTo(p-1, 0);
RotateTo(p+1, root);
int rt =ch[root][1];
ch[rt][0] = 0;
pushup(rt);
pushup(root);
}
void Reverse(int L, int R)
{
RotateTo(L-1, 0);
RotateTo(R+1, root);
f(ch[ch[root][1]][0]);
pushup(ch[root][1]);
pushup(root);
}
bool judge(int x,int y,int len)
{
uLL xval, yval;
RotateTo(x-1, 0);
if(sz[ch[root][1]] - 1 < len) return 0;
RotateTo(x+len, root);
int rt = ch[ch[root][1]][0];
xval = Hash[rt][rev[rt]];
RotateTo(y-1, 0);
if(sz[ch[root][1]] - 1 < len) return 0;
RotateTo(y+len, root);
rt = ch[ch[root][1]][0];
yval = Hash[rt][rev[rt]];
return xval == yval;
}
int Lcp(int x,int y)
{
int L = 1, R = sz[root] - 1;
if(!judge(x, y, 1)) return 0;
while(R - L > 1)
{
int mid = (L + R)>>1;
if(judge(x, y, mid)) L = mid;
else R = mid;
}
return L;
}
void Build(int l, int r,int &rt, int fa)
{
if(l > r) return;
int m = (l + r)>>1;
Newnode(rt, fa, s[m] - '0');
Build(l, m-1, ch[rt][0], rt);
Build(m+1, r, ch[rt][1], rt);
pushup(rt);
}
void Init(int n)
{
ch[0][0] = ch[0][1] = sz[0] = 0;
Hash[0][0] = Hash[0][1] = 0;
rev[0] = pre[0] = val[0] = 0;
root = tot = 0;
Newnode(root, 0, 0);
Newnode(ch[root][1], root, 0);
Build(0, n-1, ch[ch[root][1]][0], ch[root][1]);
pushup(ch[root][1]);
pushup(root);
}
int main()
{
p[0] = 1;
for(int i = 1; i < maxn; i++) p[i] = p[i-1] * pp;
int n, m;
while(scanf("%d%d",&n,&m) != EOF)
{
scanf("%s", s);
Init(n);
//debug();
int op, p1, p2, c;
for(int i = 1; i <= m; i++)
{
scanf("%d%d", &op, &p1);
if(op == 1)
{
scanf("%d", &c);
Insert(p1, c);
//debug();
}
else if(op == 2){
Delete(p1);
//debug();
}
else if(op == 3){
scanf("%d", &p2);
Reverse(p1,p2);
//debug();
}
else{
scanf("%d", &p2);
int ans = Lcp(p1, p2);
printf("%d\n", ans);
}
}
}
return 0;
}
