最長迴文

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9786 Accepted Submission(s): 3412

Problem Description

給出一個只由小寫英文字符a,b,c...y,z組成的字符串S,求S中最長迴文串的長度.
迴文就是正反讀都是一樣的字符串,如aba, abba等

Input

輸入有多組case,不超過120組,每組輸入爲一行小寫英文字符a,b,c...y,z組成的字符串S
兩組case之間由空行隔開(該空行不用處理)
字符串長度len <= 110000

Output

每一行一個整數x,對應一組case,表示該組case的字符串中所包含的最長迴文長度.

Sample Input

aaaa
abab

Sample Output

4
3

Source

2009 Multi-University Training Contest 16 - Host by NIT

Manacher大法好..O(n)

/** Author: ☆·aosaki(*’(OO)’*)  niconiconi★ **/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <list>
#include <stack>
//#include <tuple>
#define mem(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define lp(k,a) for(int k=1;k<=a;k++)
#define lp0(k,a) for(int k=0;k<a;k++)
#define lpn(k,n,a) for(int k=n;k<=a;k++)
#define lpd(k,n,a) for(int k=n;k>=a;k--)
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define lowbit(x) (x&(-x))
#define ll long long
#define pi pair<int,int>
#define vi vector<int>
#define PI acos(-1.0)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define TT cout<<"*****"<<endl;
#define TTT cout<<"********"<<endl;
inline int gcd(int a,int b)
{
    return a==0?b:gcd(b%a,a);
}

#define INF 1e9
#define eps 1e-8
#define mod 10007
#define MAX 10010

using namespace std;

const int N=200005<<1;

char T[N];   //原字符串
char S[N];   //轉換後的字符串
int  R[N];   //迴文半徑

void Init(char *T)
{
    S[0]='$';
    int len=strlen(T);
    for(int i=0;i<=len;i++)
    {
        S[2*i+1]='#';
        S[2*i+2]=T[i];
    }
}

void Manacher(char *S)
{
    int k=0,mx=0;
    int len=strlen(S);
    lp0(i,len)
    {
        if(mx>i)
            R[i]=R[2*k-i]<mx-i? R[2*k-i]:mx-i;
        else
            R[i]=1;
        while(S[i+R[i]]==S[i-R[i]])
            R[i]++;
        if(R[i]+i>mx)
        {
            mx=R[i]+i;
            k=i;
        }
    }
}

int main()
{
    int l,r;
    //freopen("in.txt","r",stdin);
    while(~scanf("%s",T))
    {
        Init(T);
        Manacher(S);
        int len=strlen(S);
        int ans=1;
        lp0(i,len)
        {
            if(ans<R[i])
            {
                ans=R[i];
                l=(i-R[i])/2; //最長迴文串左端點
                r=(i+R[i])/2-2; //右端點
            }
        }
        printf("%d\n",ans-1);
        //cout<<l<<"**"<<r<<endl;
      }
    return 0;
}