最长回文
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9097 Accepted Submission(s): 3135
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
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
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
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用manacher解决
/*************************************************************************
> File Name: hdu3068.cpp
> Author: ALex
> Mail: zchao1995@gmail.com
> Created Time: 2015年02月01日 星期日 21时27分08秒
************************************************************************/
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
const int N = 210010;
char str[N];
char mat[N];
int p[N];
void manacher (int cnt)
{
memset (p, 0, sizeof(p));
int MaxId = 0, id;
int MaxL = 0;
for (int i = 1; i < cnt; ++i)
{
if (MaxId > i)
{
p[i] = min (p[2 * id - i], MaxId - i);
}
else
{
p[i] = 1;
}
while (str[i + p[i]] == str[i - p[i]])
{
++p[i];
}
if (p[i] + i > MaxId)
{
MaxId = p[i] + i;
id = i;
}
if (p[i] - 1 > MaxL)
{
MaxL = p[i] - 1;
}
}
printf("%d\n", MaxL);
}
int main ()
{
while (~scanf("%s", mat))
{
int n = strlen (mat);
int cnt = 2;
for (int i = 0; i < n; ++i)
{
str[cnt++] = mat[i];
str[cnt++] = '#';
}
str[0] = '?';
str[1] = '#';
str[cnt] = 0;
manacher (cnt);
}
return 0;
}