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+;
char str1[N],str[N<<];
int rad[N];
char temp;
void Manacher(int *rad,char *str,int n){
int i,mx=,id;
for(int i=;i<n;i++){
if(mx>i){
rad[i]=min(rad[*id-i],mx-i);
}
else
rad[i]=;
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=;
while(~scanf("%s",str1)){
if(str1[]=='E')break;
int nn=strlen(str1);
int n=*nn+;
str[]='$';
for(int i=;i<=nn;i++){
str[*i+]='#';
str[*i+]=str1[i];
}
memset(rad,,sizeof(rad));
Manacher(rad,str,n);
int ans=;
for(int i=;i<=n;i++){
if(rad[i]>=&&rad[i]>ans){
ans=rad[i];
}
}
printf("Case %d: %d\n",cas++,ans-);
}
}

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