poj 1743 后缀数组 最长不重叠子串

Musical Theme

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 30941		Accepted: 10336

Description

A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the piano. It is unfortunate but true that this representation of melodies ignores the notion of musical timing; but, this programming task is about notes and not timings. 
Many composers structure their music around a repeating &qout;theme&qout;, which, being a subsequence of an entire melody, is a sequence of integers in our representation. A subsequence of a melody is a theme if it:

• is at least five notes long
• appears (potentially transposed -- see below) again somewhere else in the piece of music
• is disjoint from (i.e., non-overlapping with) at least one of its other appearance(s)

Transposed means that a constant positive or negative value is added to every note value in the theme subsequence. 
Given a melody, compute the length (number of notes) of the longest theme. 
One second time limit for this problem's solutions! 

Input

The input contains several test cases. The first line of each test case contains the integer N. The following n integers represent the sequence of notes. 
The last test case is followed by one zero. 

Output

For each test case, the output file should contain a single line with a single integer that represents the length of the longest theme. If there are no themes, output 0.

Sample Input

30
25 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 18
82 78 74 70 66 67 64 60 65 80
0

Sample Output

5

Hint

Use scanf instead of cin to reduce the read time.

Source

LouTiancheng@POJ

题意：

给出长度是n的数组表示音符，求最长的相同音乐段长度，相同音乐段是指两段的变化是一样的（即差值数组都一样）。

代码：

//求最长不重叠子串，把给出的数组转换成差值的形式。论文题就不多说了。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=;
const int INF=0x7fffffff;
int he[MAXN+],ra[MAXN+],sa[MAXN+],xx[MAXN+],yy[MAXN+],buc[MAXN+];
int s[MAXN+],a[MAXN+];
int len,m;
void get_suf()
{
    int *x=xx,*y=yy;
    for(int i=;i<m;i++) buc[i]=;
    for(int i=;i<len;i++) buc[x[i]=s[i]]++;
    for(int i=;i<m;i++) buc[i]+=buc[i-];
    for(int i=len-;i>=;i--) sa[--buc[x[i]]]=i;
    for(int k=;k<=len;k<<=){
        int p=;
        for(int i=len-;i>=len-k;i--) y[p++]=i;
        for(int i=;i<len;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
        for(int i=;i<m;i++) buc[i]=;
        for(int i=;i<len;i++) buc[x[y[i]]]++;
        for(int i=;i<m;i++) buc[i]+=buc[i-];
        for(int i=len-;i>=;i--) sa[--buc[x[y[i]]]]=y[i];
        swap(x,y);
        p=;x[sa[]]=;
        for(int i=;i<len;i++){
            if(y[sa[i-]]==y[sa[i]]&&y[sa[i-]+k]==y[sa[i]+k])
                x[sa[i]]=p-;
            else x[sa[i]]=p++;
        }
        if(p>=len) break;
        m=p;
    }
    for(int i=;i<len;i++) ra[sa[i]]=i;
    int k=;
    for(int i=;i<len;i++){
        if(ra[i]==) { he[]=; continue; }
        if(k) k--;
        int j=sa[ra[i]-];
        while(s[i+k]==s[j+k]&&i+k<len&&j+k<len) k++;
        he[ra[i]]=k;
    }
}
bool solve(int mid)
{
    int min_sa=INF,max_sa=-INF;
    for(int i=;i<len;i++){
        if(he[i]>=mid){
            min_sa=min(min_sa,min(sa[i],sa[i-]));
            max_sa=max(max_sa,max(sa[i],sa[i-]));
            if(min_sa+mid-<max_sa) return ;
        }else{
            min_sa=INF;max_sa=-INF;
        }
    }
    return ;
}
int main()
{
    int n;
    while(scanf("%d",&n)&&n){
        for(int i=;i<n;i++) scanf("%d",&a[i]);
        for(int i=;i<n;i++) s[i-]=a[i]-a[i-]+;
        len=n-;
        m=;
        get_suf();
        int l=,r=len,ans=;
        while(l<=r){
            int mid=(l+r)>>;
            if(solve(mid)) { ans=mid;l=mid+; }
            else r=mid-;
        }
        if(++ans<) ans=;
        printf("%d\n",ans);
    }
    return ;
}