hdu 4763 Theme Section

Theme Section

http://acm.hdu.edu.cn/showproblem.php?pid=4763

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

It's time for music! A lot of popular musicians are invited to join us in the music festival. Each of them will play one of their representative songs. To make the programs more interesting and challenging, the hosts are going to add some constraints to the rhythm of the songs, i.e., each song is required to have a 'theme section'. The theme section shall be played at the beginning, the middle, and the end of each song. More specifically, given a theme section E, the song will be in the format of 'EAEBE', where section A and section B could have arbitrary number of notes. Note that there are 26 types of notes, denoted by lower case letters 'a' - 'z'.

To get well prepared for the festival, the hosts want to know the maximum possible length of the theme section of each song. Can you help us?

 

Input

The integer N in the first line denotes the total number of songs in the festival. Each of the following N lines consists of one string, indicating the notes of the i-th (1 <= i <= N) song. The length of the string will not exceed 10^6.

 

Output

There will be N lines in the output, where the i-th line denotes the maximum possible length of the theme section of the i-th song.

 

Sample Input

5

xy

abc

aaa

aaaaba

aaxoaaaaa

 

Sample Output

0

0

1

1

2

 

Source

2013 ACM/ICPC Asia Regional Changchun Online

 

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题意：求字符串中满足 EAEBE 最长的E

（A与B可以相同，也可以没有）

 

扩展kmp求出next数组

满足第一个E和最后一个E的条件是 nxt[i]=len-nxt[i]

再在nxt[i]——i-nxt[i]范围内查找最大的nxt[]，如果>=nxt[i]，那么当前的nxt[i]就是合法的

查找区间最大值我用的线段树

#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 1000011
using namespace std;
char s[N];
int len,nxt[N];
struct node
{
    int l,r,maxn;
}tr[N*4];
void getnxt()
{
    int a=0;
    nxt[0]=len;
    while(a+1<len && s[a]==s[a+1]) a++;
    nxt[1]=a;
    int p,l,j;
    a=1;
    for(int k=2;k<len;k++)
    {
        p=a+nxt[a]-1; l=nxt[k-a];
        if(k-1+l>=p)
        {
            j=p-k+1>0 ? p-k+1 : 0;
            while(k+j<len && s[k+j]==s[j]) j++;
            nxt[k]=j;
            a=k;
          }
          else nxt[k]=l;
    }
}
void build(int k,int l,int r)
{
    tr[k].l=l; tr[k].r=r;
    if(l==r) 
    {
        tr[k].maxn=nxt[l];
        return;
    }
    int mid=l+r>>1;
    build(k<<1,l,mid);
    build(k<<1|1,mid+1,r);
    tr[k].maxn=max(tr[k<<1].maxn,tr[k<<1|1].maxn);
}
int query(int k,int opl,int opr)
{
    if(tr[k].l>=opl && tr[k].r<=opr) return tr[k].maxn;
    int mid=tr[k].l+tr[k].r>>1;
    int p=0,q=0;
    if(opl<=mid) p=query(k<<1,opl,opr);
    if(opr>mid) q=query(k<<1|1,opl,opr);
    return max(p,q); 
}
int main()
{
    int n,ans; 
    scanf("%d",&n);
    while(n--)
    {
        scanf("%s",s);
        len=strlen(s);
        getnxt();
        build(1,0,len);
        ans=0;
        for(int i=2;i<len;i++)
        {
            if(nxt[i]!=len-i) continue;
            if(i-nxt[i]<nxt[i]) continue;
            if(query(1,nxt[i],i-nxt[i])>=nxt[i]) 
            {
                ans=nxt[i];
                break;
            }
        }    
        printf("%d\n",ans);
    }
}