【HDOJ】【3068】最长回文

Manacher算法

---

　　Manacher模板题……

 //HDOJ 3068
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<iostream>
 #include<algorithm>
 #define rep(i,n) for(int i=0;i<n;++i)
 #define F(i,j,n) for(int i=j;i<=n;++i)
 #define D(i,j,n) for(int i=j;i>=n;--i)
 using namespace std;
 typedef long long LL;
 inline int getint(){
     int r=,v=; char ch=getchar();
     for(;!isdigit(ch);ch=getchar()) if(ch=='-')r=-;
     for(; isdigit(ch);ch=getchar()) v=v*+ch-'';
     return r*v;
 }
 const int N=1e5+,INF=~0u>>;
 /*******************template********************/
 char b[N];
 int p[N<<],a[N<<];
 int main(){
 #ifndef ONLINE_JUDGE
     freopen("3068.in","r",stdin);
 //    freopen("3068.out","w",stdout);
 #endif
     int n,id,mx,ans;
     while(scanf("%s",b)!=EOF){
         n=strlen(b);
         memset(p,,sizeof p);
         F(i,,n) a[i<<]=b[i-];
         n=n<<|;
         id=mx=ans=;
         F(i,,n){
             if (mx>i) p[i]=min(p[*id-i],mx-i);
             while(i-p[i]-> && i+p[i]+<=n && a[i-p[i]-]==a[i+p[i]+]) p[i]++;
             if (p[i]+i>mx) mx=p[i]+i,id=i;
             if (p[i]>ans) ans=p[i];
         }
         printf("%d\n",ans);
     }
     return ;
 }

最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9666    Accepted Submission(s): 3355

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

 

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

 

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

2009 Multi-University Training Contest 16 - Host by NIT

 

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