【POJ 3974】 Palindrome

【题目链接】

http://poj.org/problem?id=3974

【算法】

解法1 ：

字符串哈希

我们可以分别考虑奇回文子串和偶回文子串，从前往后扫描字符串，然后二分答案，检验可以用哈希

时间复杂度 ： O（TNlog(N))

解法2

Manacher算法

这个算法可以在O(n)时间内求出最长回文子串，读者可以自行查阅资料，笔者不进行详细的介绍

两种方法的效率比对

           显然，Manacher算法的复杂度是优于字符串哈希的，笔者的两份代码在POJ上的运行时间分别为4891MS和266MS

【代码】

代码1

/*
        Algorithm : Hash
        Time Complexity : O(Tnlog(n))
*/

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXL 1000010
typedef unsigned long long ULL;
const int P = ;

int i,len,l,r,mid,tmp,ans,TC;
ULL f[MAXL],ha[MAXL],hb[MAXL];
char s[MAXL];

inline ULL getha(int l,int r)
{
        return ha[r] - ha[l-] * f[r-l+];
}
inline ULL gethb(int l,int r)
{
        int tmp = l;
        l = len - r + ;
        r = len - tmp + ;
        return hb[r] - hb[l-] * f[r-l+];
}

int main()
{

        f[] = ;
        for (i = ; i < MAXL; i++) f[i] = f[i-] * P;
        while (scanf("%s",s+))
        {
                len = strlen(s+);
                if (s[] == 'E') break;
                for (i = ; i <= len; i++)    ha[i] = ha[i-] * P + (s[i] - 'a' + );
                reverse(s+,s+len+);
                for (i = ; i <= len; i++) hb[i] = hb[i-] * P + (s[i] - 'a' + );
                ans = ;
                for (i = ; i <= len; i++)
                {
                        l = ; r = min(i-,len-i);
                        tmp = ;
                        while (l <= r)
                        {
                                mid = (l + r) >> ;
                                if (getha(i-mid,i) == gethb(i,i+mid))
                                {
                                        tmp = mid;
                                        l = mid + ;
                                } else r = mid - ;
                        }
                        ans = max(ans,*tmp+);
                        l = ; r = min(i,len-i);
                        tmp = ;
                        while (l <= r)
                        {
                                mid = (l + r) >> ;
                                if (getha(i-mid+,i) == gethb(i+,i+mid))
                                {
                                        tmp = mid;
                                        l = mid + ;
                                } else r = mid - ;
                        }
                        ans = max(ans,*tmp);
                }
                printf("Case %d: %d\n",++TC,ans);
        }

        return ;

}

代码2

/*
        Algorithm : Manacher
        Time Complexity : O(TN)
*/

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 1000010

int TC;
char s[MAXN<<];

inline void Manacher()
{
        int i,len,mx = ,pos = ,ans = ;
        static char tmp[MAXN<<];
        static int p[MAXN<<];
        len = strlen(s+);
        for (i = ; i <= len; i++)
        {
                tmp[*i-] = '#';
                tmp[*i] = s[i];
        }
        tmp[len =  * len + ] = '#';
        for (i = ; i <= len; i++)
        {
                if (mx > i) p[i] = min(p[*pos-i],mx-i);
                else p[i] = ;
                while (i - p[i] >=  && i + p[i] <= len && tmp[i-p[i]] == tmp[i+p[i]]) p[i]++;
                if (i + p[i] -  > mx)
                {
                        mx = i + p[i] - ;
                        pos = i;
                }
        }
        for (i = ; i <= len; i++) ans = max(ans,p[i]-);
        printf("Case %d: %d\n",++TC,ans);
}

int main()
{

        while (scanf("%s",s+) != EOF)
        {
                if (s[] == 'E') break;
                Manacher();
        }

        return ;

}