Given a string, you are supposed to output the length of the longest symmetric sub-string. For example, given "Is PAT&TAP symmetric?", the longest symmetric sub-string is "s PAT&TAP s", hence you must output 11.
Input Specification:
Each input file contains one test case which gives a non-empty string of length no more than 1000.
Output Specification:
For each test case, simply print the maximum length in a line.
Sample Input:
Is PAT&TAP symmetric?
Sample Output:
11
思路:最长回文子串,用Manacher算法,分析见:Manacher 算法(求字符串的回文子串的最大长度)
代码:
#include <iostream>
#include <sstream>
#include <string>
#include <cstdio>
#include <queue>
#include <cstring>
#include <map>
#include <cstdlib>
#include <stack>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define mem(a,n) memset(a,n,sizeof(a))
#define DBGS() cout<<"START\n"
#define DBGE() cout<<"END\n"
const int N = 1e5+5;
const ll INF = 0x3f3f3f3f;
const double eps=1e-4;
int p[N];
char s[N],t[N];
int init() {
int len=strlen(s);
t[0]='$',t[1]='#';
int j=2;
for(int i=0; i<len; i++) {
t[j++]=s[i];
t[j++]='#';
}
s[j]='\0';
return j;
}
void manacher() {
int len=init();
int maxlen=0,mx=0,id;
for(int i=1; i<len; i++) {
if(i<mx)
p[i]=min(p[id*2-i],mx-i);
else
p[i]=1;
while(t[i-p[i]]==t[i+p[i]])
p[i]++;
if(mx<i+p[i]) {
id=i;
mx=i+p[i];
}
maxlen=max(maxlen,p[i]-1);
}
printf("%d\n",maxlen);
}
int main() {
fgets(s,sizeof(s),stdin);
manacher();
return 0;
}