poj 1743 Musical Theme 后缀数组

Musical Theme

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 15246 Accepted: 5267

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

3025 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 1882 78 74 70 66 67 64 60 65 800

Sample Output

5

Hint

Use scanf instead of cin to reduce the read time.

Source

LouTiancheng@POJ

2009年罗穗骞的论文处理字符串的有力工具---后缀数组

例题，经典中的经典

以下摘自论文内容：

不可重叠最长重复子串
给定一个字符串，求最长重复子串，这两个子串不能重叠。

先二分答案，把题目变成判定性问题：判断是否
存在两个长度为k 的子串是相同的，且不重叠。解决这个问题的关键还是利用height 数组。把排序后的后缀分成若干组，其中每组的后缀之间的height 值都不小于k。例如，字符串为“aabaaaab”，当k=2 时，后缀分成了4 组，如图5所示。

容易看出，有希望成为最长公共前缀不小于k 的两个后缀一定在同一组。然后对于每组后缀，只须判断每个后缀的sa 值的最大值和最小值之差是否不小于k。如果有一组满足，则说明存在，否则不存在。整个做法的时间复杂度为O(nlogn)。

倍增算法：

#include <stdio.h>#define maxn 20000int wa[maxn],wb[maxn],wv[maxn],ws[maxn];int cmp(int *r,int a,int b,int l){return r[a]==r[b]&&r[a+l]==r[b+l];}void da(int *r,int *sa,int n,int m){     int i,j,p,*x=wa,*y=wb,*t;     for(i=0;i<m;i++) ws[i]=0;     for(i=0;i<n;i++) ws[x[i]=r[i]]++;     for(i=1;i<m;i++) ws[i]+=ws[i-1];     for(i=n-1;i>=0;i--) sa[--ws[x[i]]]=i;     for(j=1,p=1;p<n;j*=2,m=p)     {       for(p=0,i=n-j;i<n;i++) y[p++]=i;       for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;       for(i=0;i<n;i++) wv[i]=x[y[i]];       for(i=0;i<m;i++) ws[i]=0;       for(i=0;i<n;i++) ws[wv[i]]++;       for(i=1;i<m;i++) ws[i]+=ws[i-1];       for(i=n-1;i>=0;i--) sa[--ws[wv[i]]]=y[i];       for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)       x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;     }     return;}int rank[maxn],height[maxn];void calheight(int *r,int *sa,int n){     int i,j,k=0;     for(i=1;i<=n;i++) rank[sa[i]]=i;     for(i=0;i<n;height[rank[i++]]=k)     for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);     return;}int check(int *sa,int n,int k){    int i,max=sa[1],min=sa[1];    for(i=2;i<=n;i++)    {      if(height[i]<k) max=min=sa[i];      else      {        if(sa[i]<min) min=sa[i];        if(sa[i]>max) max=sa[i];        if(max-min>k) return(1);      }    }    return(0);}int r[maxn],sa[maxn];int main(){    int i,j=0,k,n;    int min,mid,max;    scanf("%d",&n);    while(n!=0)    {      n--;      scanf("%d",&j);      for(i=0;i<n;i++)      {        scanf("%d",&k);        r[i]=k-j+100;        j=k;      }      r[n]=0;      da(r,sa,n+1,200);      calheight(r,sa,n);      min=1;max=n/2;      while(min<=max)      {        mid=(min+max)/2;        if(check(sa,n,mid)) min=mid+1;        else max=mid-1;      }      if(max>=4) printf("%d\n",max+1);      else printf("0\n");      scanf("%d",&n);    }    return 0;}

DC3算法：

#include <stdio.h>#define maxn 20002#define F(x) ((x)/3+((x)%3==1?0:tb))#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)int wa[maxn],wb[maxn],wv[maxn],ws[maxn];int c0(int *r,int a,int b){return r[a]==r[b]&&r[a+1]==r[b+1]&&r[a+2]==r[b+2];}int c12(int k,int *r,int a,int b){if(k==2) return r[a]<r[b]||r[a]==r[b]&&c12(1,r,a+1,b+1); else return r[a]<r[b]||r[a]==r[b]&&wv[a+1]<wv[b+1];}void sort(int *r,int *a,int *b,int n,int m){     int i;     for(i=0;i<n;i++) wv[i]=r[a[i]];     for(i=0;i<m;i++) ws[i]=0;     for(i=0;i<n;i++) ws[wv[i]]++;     for(i=1;i<m;i++) ws[i]+=ws[i-1];     for(i=n-1;i>=0;i--) b[--ws[wv[i]]]=a[i];     return;}void dc3(int *r,int *sa,int n,int m){     int i,j,*rn=r+n,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p;     r[n]=r[n+1]=0;     for(i=0;i<n;i++) if(i%3!=0) wa[tbc++]=i;     sort(r+2,wa,wb,tbc,m);     sort(r+1,wb,wa,tbc,m);     sort(r,wa,wb,tbc,m);     for(p=1,rn[F(wb[0])]=0,i=1;i<tbc;i++)     rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++;     if(p<tbc) dc3(rn,san,tbc,p);     else for(i=0;i<tbc;i++) san[rn[i]]=i;     for(i=0;i<tbc;i++) if(san[i]<tb) wb[ta++]=san[i]*3;     if(n%3==1) wb[ta++]=n-1;     sort(r,wb,wa,ta,m);     for(i=0;i<tbc;i++) wv[wb[i]=G(san[i])]=i;     for(i=0,j=0,p=0;i<ta && j<tbc;p++)     sa[p]=c12(wb[j]%3,r,wa[i],wb[j])?wa[i++]:wb[j++];     for(;i<ta;p++) sa[p]=wa[i++];     for(;j<tbc;p++) sa[p]=wb[j++];     return;}int rank[maxn],height[maxn];void calheight(int *r,int *sa,int n){     int i,j,k=0;     for(i=1;i<=n;i++) rank[sa[i]]=i;     for(i=0;i<n;height[rank[i++]]=k)     for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);     return;}int check(int *sa,int n,int k){    int i,max=sa[1],min=sa[1];    for(i=2;i<=n;i++)    {      if(height[i]<k) max=min=sa[i];      else      {        if(sa[i]<min) min=sa[i];        if(sa[i]>max) max=sa[i];        if(max-min>k) return(1);      }    }    return(0);}int r[maxn*3],sa[maxn*3];int main(){    int i,j=0,k,n;    int min,mid,max;    scanf("%d",&n);    while(n!=0)    {      n--;      scanf("%d",&j);      for(i=0;i<n;i++)      {        scanf("%d",&k);        r[i]=k-j+100;        j=k;      }      r[n]=0;      dc3(r,sa,n+1,200);      calheight(r,sa,n);      min=1;max=n/2;      while(min<=max)      {        mid=(min+max)/2;        if(check(sa,n,mid)) min=mid+1;        else max=mid-1;      }      if(max>=4) printf("%d\n",max+1);      else printf("0\n");      scanf("%d",&n);    }    return 0;}