poj 1743

Musical Theme
Time Limit: 1000MS        Memory Limit: 30000K
Total Submissions: 31563        Accepted: 10523
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

问你最长连续不重合子串

#define  LL long long
using namespace std;
const int MAXN=20010;
const int maxn=20010;
int 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)//用倍增求后缀数组
{
      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]]++;//以字符的ascii码为下标
      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 sa[MAXN],Rank[MAXN],height[MAXN];
//求height数组
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++);
      // Unified
      for(int i=n;i>=1;--i) ++sa[i],Rank[i]=Rank[i-1];
}
int check(int n,int k){
    int maxx=sa[1],minx=sa[1];
    for(int i=2;i<=n;++i){
        if(height[i]<k) maxx=minx=sa[i];
        else{
            if(sa[i]<minx) minx=sa[i];
            if(sa[i]>maxx) maxx=sa[i];
            if(maxx-minx>=k) return true;
        }//每个后缀sa的最大值和最小值只差是否不小于k；
    }
    return false;
}
int str[maxn],n;
int main()
{

    while(~scanf("%d",&n)&&n)
    {
            for(int i=0;i<n;++i)
            scanf("%d",&str[i]);
        for(int i=n-1;i>0;--i)
            str[i]=str[i]-str[i-1]+88;
        for(int i=0;i<n-1;++i)
            str[i]=str[i+1];
          str[n-1]=0;
      da(str,sa,n,176);
      calheight(str,sa,n-1);

       int l=4,r=(n-1)/2;
        int ans=0;
        while(l<=r){
            int mid=(l+r)>>1;
            if(check(n-1,mid)){
                l=mid+1;
                ans=max(mid,ans);
            }
            else
                r=mid-1;
        }//二分搜索存不存在
        if(ans==0) printf("0
");
        else printf("%d
",ans+1);
    }
    return 0;
}