Palindrome

poj3974:http://poj.org/problem?id=3974

题意：求给定长度最长回文串的长度。

题解：直接套manacher,搞定。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=2e6+440;
char str1[N],str[N<<1];
int rad[N];
char temp;
void Manacher(int *rad,char *str,int n){
   int i,mx=0,id;
   for(int i=1;i<n;i++){
    if(mx>i){
    rad[i]=min(rad[2*id-i],mx-i);
   }
   else
    rad[i]=1;
    for(;str[i-rad[i]]==str[i+rad[i]];rad[i]++){
        if(rad[i]+i>mx){
            mx=rad[i]+i;
            id=i;
        }
     }
   }
}

int main(){
    int cas=1;
  while(~scanf("%s",str1)){
       if(str1[0]=='E')break;
      int nn=strlen(str1);
        int n=2*nn+2;
        str[0]='$';
      for(int i=0;i<=nn;i++){
          str[2*i+1]='#';
          str[2*i+2]=str1[i];
      }
      memset(rad,0,sizeof(rad));
      Manacher(rad,str,n);
      int ans=0;
      for(int i=0;i<=n;i++){
        if(rad[i]>=3&&rad[i]>ans){
            ans=rad[i];
        }
      }
      printf("Case %d: %d
",cas++,ans-1);
  }
}