manacher算法模板程序 HDOJ3068

最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23931    Accepted Submission(s): 8761

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

aaaaabab

 

Sample Output

43

 

Source

2009 Multi-University Training Contest 16 - Host by NIT

 

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直接贴代码吧

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>using namespace std;const int maxn = 110000+5;char s[maxn],s1[maxn<<1];int a[maxn<<1];int i,j,k,max_right,pos,len,new_len;void init(){//构建新的字符串    len = strlen(s);    pos = 0;    new_len = 0;    s1[0]='*';    for (i=0; i<len; i++){        s1[++new_len] = '#';        s1[++new_len] = s[i];    }    s1[++new_len] = '#';}int manacher(){    int MAX = -1;    for (i=2; i<2*len+1; i++) {        if (a[pos]+pos>i) a[i] = min(a[pos*2-i],a[pos]+pos-i);        else a[i] = 1;        while (s1[i-a[i]]==s1[i+a[i]]) a[i]++;        if (pos+a[pos]<i+a[i]) pos = i;        MAX = max(MAX,a[i]);    }    return MAX-1;}int main(){    while (scanf("%s",s)!=EOF){        init();        printf("%d\n",manacher());    }    return 0;}