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。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 const int N=2e6+440; 7 char str1[N],str[N<<1]; 8 int rad[N]; 9 char temp; 10 void Manacher(int *rad,char *str,int n){ 11 int i,mx=0,id; 12 for(int i=1;i<n;i++){ 13 if(mx>i){ 14 rad[i]=min(rad[2*id-i],mx-i); 15 } 16 else 17 rad[i]=1; 18 for(;str[i-rad[i]]==str[i+rad[i]];rad[i]++){ 19 if(rad[i]+i>mx){ 20 mx=rad[i]+i; 21 id=i; 22 } 23 } 24 } 25 } 26 27 int main(){ 28 int cas=1; 29 while(~scanf("%s",str1)){ 30 if(str1[0]=='E')break; 31 int nn=strlen(str1); 32 int n=2*nn+2; 33 str[0]='$'; 34 for(int i=0;i<=nn;i++){ 35 str[2*i+1]='#'; 36 str[2*i+2]=str1[i]; 37 } 38 memset(rad,0,sizeof(rad)); 39 Manacher(rad,str,n); 40 int ans=0; 41 for(int i=0;i<=n;i++){ 42 if(rad[i]>=3&&rad[i]>ans){ 43 ans=rad[i]; 44 } 45 } 46 printf("Case %d: %d ",cas++,ans-1); 47 } 48 }