hihocoder 1032 manachar 求回文串O(n)

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cmath>
#include <cstring>
#include <stack>
#include <set>
#include <map>
#include <vector>

using namespace std;
#define INF 0x2fffffff
#define LL long long
#define MAX(a,b) ((a)>(b))?(a):(b)
#define MIN(a,b) ((a)<(b))?(a):(b)
char a[1000005];
char b[2000005];
int  p[2000005];
int newstring(){
    b[0] = '$';
    b[1] = '#';
    int len = strlen(a);
    int j = 2;
    for(int i = 0;i < len;i++){
        b[j] = a[i];
        b[++j] = '#';
        ++j;    
    }
    b[j] = '';
} 
int manachar(){
    int len = strlen(b);
    int id = 0;
    int maxd = 1;
    int ans = 0;
    for(int i = 1;i < len;i++){
        if(maxd > i){
            p[i] = min(p[2*id-i],maxd-i);
        }else{
            p[i] = 1;
        }
        while(b[p[i]+i] == b[i-p[i]]) 
            p[i]++;
        if(p[i]+i > maxd){
            maxd = p[i]+i;
            id = i;
        }
        if(ans < p[i]){
            ans = p[i];
        } 
    }
    printf("%d
",ans-1);
}
int main(){
    int t;
    cin >> t;
    while(t--){
        scanf("%s",a);
        newstring();
        manachar();
    }   
    return 0;
}

依据回文串左右是全然同样的性质。左边求出来后非常快的就能求出来右边的长度