hdu 3068 最长回文 扩展KMP

最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1301    Accepted Submission(s): 313

Problem Description

给出一个只由小写英文字符a,b,c...y,z组成的字符串S,求S中最长回文串的长度.
回文就是正反读都是一样的字符串,如aba, abba等

 

Input

输入有多组case,不超过120组,每组输入为一行小写英文字符a,b,c...y,z组成的字符串S
两组case之间由空行隔开(该空行不用处理)
字符串长度len <= 110000

 

Output

每一行一个整数x,对应一组case,表示该组case的字符串中所包含的最长回文长度.

 

Sample Input

aaaa
abab

 

Sample Output

4
3

 

Source

2009 Multi-University Training Contest 16 - Host by NIT

 

Recommend

lcy

 

 

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
using namespace std;

#define max_size 110010

int next1[max_size];
int next2[max_size];
int next[max_size];
int ans;

char a[max_size];
char b[max_size];

void getNext1(int n)
{
     int i, j, k;
     next[0] = n;
     j = 0;
     while (j + 1 < n && b[j] == b[j+1])j++;
     next[1] = j;
     k = 1;
     for (i = 2; i < n; i++)
     {
         if (next[i-k] + i < next[k] + k)
            next[i] = next[i-k];
         else
         {
             j = next[k] + k - i;
             if (j < 0)j = 0;
             while (i+j < n && b[j] == b[i+j])j++;
             next[i] = j;
             k = i;
         }
     }
}

void getNext2(char *s, int *nexts, int n, int m)
{
     getNext1(m);
     int i, j, k;
     j = 0;
     k = 0;
     while (s[j] == b[j])j++;
     nexts[0] = j;
     
     for (i = 1; i < n; i++)
     {
         if (next[i-k] + i < nexts[k] + k)
             nexts[i] = next[i-k];
         else
         {
             j = nexts[k] + k - i;
             if (j < 0) j = 0;
             while (i + j < n && s[i+j] == b[j])j++;
             nexts[i] = j;
             k = i;
         }
     }
}

void res(char *s, int n)
{
     char st;
     int len = (n>>1);
     for (int i = 0; i < len; i++)
     {
         st = s[i];
         s[i] = s[n-i-1];
         s[n-i-1] = st;
     }
}

void find(char *s, int n)
{
     if (ans >= n || n < 2)return;
     int i, k, x;
     int mid;
     
     mid = (n>>1);
     
     for (i = mid; i < n; i++)b[i-mid] = s[i];
     b[i-mid] = 0;
     res(s, n);
     getNext2(s, next1, n, i-mid);
     
     res(s, n);
     for (i = 0; i < mid; i++)
       b[i] = s[mid-i-1];
     b[i] = 0;
     getNext2(s, next2, n, mid);
     
     next1[n] = next2[n] = 0;
     
     for (i = 0; i < mid; i++)
     {
         if (next2[i] >= mid - i)
         {
             x = mid - i + 2 * next1[n-i];
             if (ans < x)ans = x;
         }
     }
     for (i = mid; i < n; i++)
     {
         if (next1[n-i] >= i - mid)
         {
             x = i - mid + 2*next2[i];
             if (ans < x) ans = x;
         }
     }
     find(s, mid);
     find(s+mid, n-mid);
}

int main()
{
    int  n;
    while (scanf("%s", a) != EOF)
    {
          n = strlen(a);
          ans = 1;
          find(a, n);
          printf("%d\n", ans);
    }
    return 0;
}