HDU 3068 最长回文 【最长回文子串】

和上一题一样，不过这题只是要求最长回文子串的长度

在此采用了非常好用的Manacher算法

据说还是O(n) 的效率QAQ 

详细用法参考了上篇博客的参考资料，这两天有空学习一下~

Source code:

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0)

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const double eps = 1e-7      ;
const int N = 1              ;
const int M = 1100011*2      ;
const ll P = 10000000097ll   ;

char str[M];//start from index 1
int p[M];
char s[M];
int n;

void kp(){
    int i;
    int mx = 0;
    int id;
    for(i = 1; i < n; ++i){
        if( mx > i )
            p[i] = Min( p[2*id-i], p[id]+id-i );
        else
            p[i] = 1;
        for(; str[i+p[i]] == str[i-p[i]]; p[i]++)
            ;
        if( p[i] + i > mx ){
            mx = p[i] + i;
            id = i;
        }
    }
}

void pre(){
    int i,j,k;
    n = strlen(s);
    str[0] = '$';
    str[1] = '#';
    for(i = 0; i < n; ++i){
        str[i*2 + 2] = s[i];
        str[i*2 + 3] = '#';
    }
    n = n*2 + 2;
    str[n] = 0;
}

int main(){
    int T,_=0, t, i, ans;
    while(EOF != scanf("%s", s)){
        ans = 0;
        pre();
        kp();
        for(i = 0; i < n; ++i)
            checkmax(ans, p[i]);
        printf("%d
", ans-1);
    }
    return 0;
}