Codeforces Gym 100570 E. Palindrome Query Manacher

E. Palindrome Query
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100570/problem/E

Description

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. 1 p x : Modify sp = x where 1 ≤ p ≤ n and x is a lower case English letter.

2. 2 p : 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. 3 p : 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.

Input

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.

Output

For each query of type 2 and 3 print the answer in a single line.

Sample Input

abcd 3
3 1
1 2 c
3 2

Sample Output

-1
2

HINT

 

题意

给你一个字符串，然后有3个操作

1.修改一个位置的字符为

2.查询以p为中心的奇数回文串最长长度

3.查询以p为中心的偶数回文串最长长度

题解：

这道题看起来很唬人，其实暴力可过……

直接傻逼暴力就好

我拍的是manacher，直接暴力修改，也是直接过了……

代码

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 201001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************
char s[maxn];
char str[maxn];
int p[maxn];
int dp1[maxn];
int dp2[maxn];
int l;
void manacher(char s[],int l)
{
   int i,j,k,ans=0;
   for(i=1;i<=l;++i)str[i<<1]=s[i],str[(i<<1)+1]='#';
   str[1]='#';str[l*2+1]='#';str[0]='&';str[l*2+2]='$';
   l=l*2+1;j=0;
   for(i=1;i<=l;)
   {
       while(str[i-j-1]==str[i+j+1])++j;
       p[i]=j;if(j>ans)ans=j;
       for(k=1;k<=j&&p[i]-k!=p[i-k];++k)p[i+k]=min(p[i-k],p[i]-k);
       i+=k;j=max(j-k,0);
   }
}
struct node
{
    int p;
    char c;
};
vector<node> Q;
int main()
{
    //test;
    int m;
    scanf("%s",s+1);
    scanf("%d",&m);
    l=strlen(s+1);
    manacher(s,l);
    l=l*2+1;
    for(int ii=0;ii<m;ii++)
    {
        int k;
        scanf("%d",&k);
        if(k==1)
        {
            int pp;
            pp=read();
            char c;
            scanf("%c",&c);
            Q.push_back((node){pp,c});
        }
        if(k==2)
        {
            if(Q.size())
            {
                for(int i=0;i<Q.size();i++)
                {
                    s[Q[i].p]=Q[i].c;
                }
                Q.clear();
                manacher(s,l/2);
            }
            int pp=read();
            printf("%d\n",p[pp*2]);
        }
        if(k==3)
        {
            if(Q.size())
            {
                for(int i=0;i<Q.size();i++)
                {
                    s[Q[i].p]=Q[i].c;
                }
                Q.clear();
                manacher(s,l/2);
            }
            int pp=read();
            if(p[pp*2+1]==0)
                printf("-1\n");
            else
                printf("%d\n",p[pp*2+1]);
        }
    }
}