HDU 3068 最长回文

来源：互联网发布：网络戴口罩女主播照片编辑：程序博客网时间：2024/06/18 16:52

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

分析

马拉车模板题

代码

#include <algorithm>#include <iostream>#include <cstring>#include <complex>#include <cstdio>#include <queue>#include <cmath>#include <map>#include <set>#define N 110005#define INF 0x7fffffff#define sqr(x) ((x) * (x))#define pi acos(-1)char ch[N << 1];int len[N << 1];int read(){    int x = 0, f = 1;    char ch = getchar();    while (ch < '0' || ch > '9'){if (ch == '-') f = -1; ch = getchar();}    while (ch >= '0' && ch <= '9')  {x = x * 10 + ch - '0'; ch = getchar();}    return x * f;}void manacher(char s[], int l){    int L = 0;    ch[L++] = '$';    ch[L++] = '#';    for (int i = 0; i < l; i++)        ch[L++] = s[i], ch[L++] = '#';    len[L] = 0;    int mx = 0, p = 0;    for (int i = 0; i < L; i++)    {        len[i] = mx > i ? std::min(len[2 * p - i], mx - i) : 1;        while (ch[i + len[i]] == ch[i - len[i]]) len[i]++;        if (i + len[i] > mx)            mx = i + len[i], p = i;    }}char s[N];int main(){    while(~scanf("%s",s))      {          int L = strlen(s);          manacher(s, L);          int ans = 0;          for (int i = 0; i < 2 * L + 2; i++)          {              ans = std::max(ans, len[i] - 1);          }          printf("%d\n", ans);      }  }

0 0