poj3974 Palindrome

题目连接 

题意：求一个字符串的最长回文子串

题解：是一个Manacher模板题，为了统一奇偶，先预处理在字符间添加间隔，使字符串长度变为偶数，例如"abc"添加分隔符后变成"$#a#b#c#"，剩下的就很简单了，这个题也可以用哈希做。就时间复杂度来看Manacher明显比哈希快。

Manacher代码

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<string>
using namespace std;

int p[2000020];//p数组需要开至少字符串两倍的空间
char s[1010100];

int Manacher(char str[])
{
    memset(p,0,sizeof(p));
    string s="$#";
    int len=strlen(str);
    for(int i=0;i<len;i++){
        s+=str[i];
        s+='#';
    }
    int mx=0,id=0,reslen=0,rescenter=0;
    len=s.length();
    for(int i=1;i<=len;i++)
    {
        p[i]=mx>i?min(p[2*id-i],mx-i):1;//
        while(s[i+p[i]]==s[i-p[i]])p[i]++;
        if(mx<i+p[i])
        {
            mx=i+p[i];
            id=i;
        }
        if(reslen<p[i])
        {
            reslen=p[i];
            rescenter=i;
        }
    }
    return reslen-1;
}


int main()
{
    int t=1;
    while(~scanf("%s",s)){
        if(strcmp(s,"END")==0) break;
        printf("Case %d: %d
",t++,Manacher(s));
    }
    return 0;
}

哈希代码

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;

const int maxn = 2e6+20;
const int base = 131;　　　　　//不易冲突
typedef unsigned long long ULL;
ULL f[maxn], f2[maxn], t[maxn];

char s[maxn];
ULL solve(ULL h[],int l, int r)
{
    return h[r] - h[l-1] * t[r-l+1];
}

int main()
{
    int T = 1;
    while(scanf("%s", s+1) != EOF) {
        int len = strlen(s+1);
        if(!strcmp(s+1, "END"))
            break;
        for(int i = 2*len; i > 0; i-=2) {
            s[i] = s[i/2];
            s[i-1] = 'z'+1;
        }
        len *= 2;
        f[0] = f2[0] = 0;
        t[0] = 1;
        for(int i = 1, j = len; i <= len; i++, j--) {
            f[i] = f[i-1]*base + (s[i] - 'a' + 1);
            f2[i] = f2[i-1]*base + (s[j] - 'a' + 1);
            t[i] = t[i-1] * base;
        }
        int l, r;
        int ans = 0;
        int mid;
        for(int i = 1; i <= len; i++) {
            l = 0, r = min(i-1, len-i);
            while(l < r) {
                mid = (l+r+1)>>1;
                if(solve(f,i-mid,i-1) != solve(f2,len-(i+mid)+1,len-(i+1)+1))
                    r = mid-1;
                else l = mid;
            }
            if(s[i-l] <= 'z')
                ans = max(ans, l+1);
            else
                ans = max(ans, l);
        }
        printf("Case %d: %d
", T++, ans);
    }
    return 0;
}