De Prezer loves palindrome strings. A string s1s2...sn is palindrome if and only if it is equal to its reverse.
De Prezer also loves queries.
You are given string s of length n and m queries. There are 3 types of queries :
1. 1px : Modify sp = x where 1 ≤ p ≤ n and x is a lower case English letter.
2. 2p : Print the length of the largest palindrome substring of s like slsl + 1...sr such that l ≤ p ≤ r and r - p = p - l. (1 ≤ p ≤ n)
3. 3p : Print the length of the largest palindrome substring of s like slsl + 1...sr such that l ≤ p and p + 1 ≤ r and r - p - 1 = p - l. (1 ≤ p ≤ n - 1) or - 1 if there is no such substring.
The first line of input contains s and m.
Next m lines contain queries.
1 ≤ n, m ≤ 105
s only contains lower case English letters.
For each query of type 2 and 3 print the answer in a single line.
解题报告：
大意就是给你一个字符串，有三种操作:
操作1：将某个位置的字符改成另外一个字符
操作2：询问以位置 x 位中点的奇回文串长度
操作3：询问以位置 x,x+1位中的偶回文串长度
我们建立一颗线段树来解决这个问题
线段树中的存储的 key1 表示 s[L] * p^0 + S[L+1] * p ^1 + ..... S[R] * p ^ (R-L) , key2 值表示S[R] * p ^ 0 + S[R-1] * p ^1 + ...S[L] * p^(R-L)
显然key2 是倒着的，这样方便我们查询这个区间是否是回文子串
题目显然要求是单点更新，区间查询.
对于单点更新而言，最重要的就是push_up函数，我们考虑
将[L,mid] 和 [mid + 1 , R]进行合并时，注意到key值的含义，我们需要给右边的[mid+1,R]的key1值乘上p^(mid-L+1),key2则是给[L,mid]乘上(R-mid+1)
那么更新我们就解决了,查询呢
下面皆以 key1 为例(key2同理可得)
对于key1而言，我们主要就是要给落在右边的查询区间段上的乘上一个值.
假设我们现在要查询的区间段是[ ql , qr ] ,此时所在的线段是[L,R],中点是mid , 我们假设 qr > mid(即有落在右边的部分)
那么我们需要给右边乘的值就是p ^ (mid - ql + 1) <仔细想想>
key2同理可得，这里不再累述
这样我们就解决了查询问题，那么对于每个询问，我们二分答案即可
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <iomanip>
#include <string>
#include <ctime>
typedef unsigned char byte;
#define pb push_back
#define input_fast std::ios::sync_with_stdio(false);std::cin.tie(0)
#define local freopen("in.txt","r",stdin)
#define pi acos(-1)
using namespace std;
typedef pair<unsigned int ,unsigned int> dl;
const int maxn = 1e5 + 500;
const unsigned int p1 = 131;
const unsigned int p2 = 171;
unsigned int value1[maxn];
unsigned int value2[maxn];
typedef struct treenode
{
int l , r ;
dl key1,key2;
void updata(unsigned int v)
{
key1.first = key1.second = v;
key2.first = key2.second = v;
}
};
treenode tree[maxn * 4];
inline void build_tree(int o,int l,int r)
{
tree[o].l = l , tree[o].r = r , tree[o].key1.first = tree[o].key1.second = tree[o].key2.first = tree[o].key2.second = 0;
if (r > l)
{
int mid = l + (r-l)/2;
build_tree(2*o,l,mid);
build_tree(2*o+1,mid+1,r);
}
}
inline void push_up(int o)
{
int ac = tree[2*o].r - tree[2*o].l + 1 ;
int ac2 = tree[2*o+1].r - tree[2*o+1].l + 1;
tree[o].key1.first = tree[2*o].key1.first + tree[2*o+1].key1.first * value1[ac];
tree[o].key1.second = tree[2*o].key1.second + tree[2*o+1].key1.second * value2[ac];
tree[o].key2.first = tree[2*o].key2.first * value1[ac2] + tree[2*o+1].key2.first;
tree[o].key2.second = tree[2*o].key2.second * value2[ac2] + tree[2*o+1].key2.second;
}
void updata(int ql,int qr,int o,unsigned int v)
{
int l = tree[o].l , r = tree[o].r;
if (ql <= l && qr >= r)
tree[o].updata(v);
else
{
int mid = l + (r-l) / 2;
if (mid >= ql)
updata(ql,qr,2*o,v);
if (mid < qr)
updata(ql,qr,2*o+1,v);
push_up(o);
}
}
dl query(int ql,int qr,int o,int type)
{
int l = tree[o].l , r = tree[o].r;
//cout << "L is " << l << " R is " << r << " ql is " << ql << " qr is " << qr << endl;
if (ql == l && qr == r)
{
if (type == 1) return tree[o].key1;
else return tree[o].key2;
}
else
{
int mid = l + (r-l) / 2 , ac;
dl res(0,0) , temp;
if (mid >= ql)
{
int atr = min(mid,qr);
temp = query(ql,atr,2*o,type);
if (type == 2 && qr - mid > 0)
{
temp.first *= value1[qr - mid];
temp.second *= value2[qr - mid];
}
res.first += temp.first;
res.second += temp.second;
}
if (mid < qr)
{
int ltr = max(ql,mid+1);
temp = query(ltr,qr,2*o+1,type);
if (type == 1 && mid - ql + 1 > 0)
{
temp.first *= value1[mid - ql + 1];
temp.second *= value2[mid - ql + 1];
}
res.first += temp.first;
res.second += temp.second;
}
//cout << "L is " << l << " R is " << r << " fist-value is " << res.first << " second-value is " << res.second << endl;
return res;
}
}
char str[maxn] , temp[200];
int length , m ;
bool equaldl(dl x,dl y)
{
return x.first == y.first && x.second == y.second;
}
int main(int argc,char *argv[])
{
scanf("%s%d",str,&m);
value1[0] = value2[0] = 1;
length = strlen(str);
build_tree(1,0,length-1);
for(int i = 1 ; i <= length ; ++ i)
{
value1[i] = value1[i-1]*p1;
value2[i] = value2[i-1]*p2;
}
for(int i = 0 ; i < length ; ++ i) updata(i,i,1,(unsigned int)str[i]);
while(m--)
{
int type,x;
scanf("%d",&type);
if (type == 1)
{
scanf("%d%s",&x,temp);
x--;
str[x] = temp[0];
updata(x,x,1,temp[0]);
}
else
{
scanf("%d",&x);
int L , R;
x--;
if (type == 2)
{
L = 1 , R = min( x + 1 , length - x );
while(L < R)
{
int mid = L + (R-L+1) / 2;
int tl = x - mid + 1 ;
int tr = x + mid - 1 ;
if (equaldl(query(tl,tr,1,1),query(tl,tr,1,2))) L = mid;
else R = mid - 1;
}
printf("%d\n",2*L-1);
}
else
{
if (x+1 >= length || str[x] != str[x+1]) printf("-1\n");
else
{
int L = 1 , R = min(x + 1 , length - x - 1);
while(L < R)
{
int mid = L + (R-L+1) / 2;
int tl = x - mid + 1;
int tr = x + mid;
if (equaldl(query(tl,tr,1,1),query(tl,tr,1,2))) L = mid;
else R = mid - 1;
}
printf("%d\n",2*L);
}
}
}
}
return 0;
}