POJ-1743

Musical Theme

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 21134		Accepted: 7234

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：百度的题意，根据罗的论文知道了题意，但是没有懂，百度了一下，
                    给出一段只有音高(整数表示),没有节奏的乐谱,问其中最长的曲调相同的没有重叠的两段的长度是多少。
          做法：将给的高音进行后以为 - 前一位 + 90 ,然后进行da(),并且用二分进行查
                    
**/
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <queue>
#define maxn 20000 +100
using namespace std;
int t1[maxn],t2[maxn],c[maxn];
bool cmp(int *r,int a,int b,int l)
{
    return r[a] == r[b] && r[a+l] == r[b+l];
}
int sa[maxn];
int Rank[maxn];
int heigth[maxn];
void da(int str[],int n,int m)
{
    n++;
    int i,j,p, *x = t1,*y = t2;
    for(i=0; i<m; i++) c[i] = 0;
    for(i=0; i<n; i++) c[x[i] = str[i]]++;
    for(i=1; i<m; i++) c[i] += c[i-1];
    for(i=n-1; i>=0; i--) sa[--c[x[i]]] = i;
    for(j=1; j<=n; j<<=1)
    {
        p = 0;
        for(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<m; i++) c[i] = 0;
        for(i=0; i<n; i++) c[x[y[i]]]++;
        for(i=1; i<m; i++) c[i] += c[i-1];
        for(i=n-1; i>=0; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x,y);
        p = 1;
        x[sa[0]] = 0;
        for(i=1; i<n; i++)
            x[sa[i]] = cmp(y,sa[i-1],sa[i],j)?p-1:p++;
        if(p >=n) break;
        m = p;
    }
    int k =0 ;
    n--;
    for(i=0; i<=n; i++) Rank[sa[i]] = i;
    for(i =0 ; i<n; heigth[Rank[i++]] = k)
    {
        for(k?k--:0,j = sa[Rank[i]-1]; str[i+k] == str[j+k]; k++);
    }
}
int n;
int mm[maxn];
bool check(int mid,int len)
{
    for(int i=2; i<=len; i++)
    {
        if(heigth[i] < mid) continue;  ///因为heigth数组中存储的是suffix(sa[i])和suffix(sa[i-1])的公共前缀的个数
                                                                                /// 所以只有但heigth[i] >= mid 才能进行查询 否则没有
        for(int  j= i-1; j>=2; j--)
        {
            if(abs(sa[i] - sa[j]) >= mid)                     
            {
                return true;
            }
            if(heigth[j] < mid) break;
        }
    }
    return false;
}
int main()
{
//#ifndef ONLINE_JUDGE
//    freopen("in.txt","r",stdin);
//#endif
    while(~scanf("%d",&n))
    {
        if(n == 0) break;
        memset(heigth,0,sizeof(heigth));
        memset(mm,0,sizeof(mm));
        memset(Rank,0,sizeof(Rank));
        memset(sa,0,sizeof(sa));
        n--;
        for(int i=0; i<=n; i++)
        {
            scanf("%d",&mm[i]);
        }
        if(n < 10)
        {
            printf("0\n");
            continue;
        }
        for(int i=0; i<n; i++)
        {
            mm[i] = mm[i+1] - mm[i] +90;
        }
        mm[n] = 0;
        da(mm,n,200);
        int left = 3,mid,right = n;
        int ans = 0;
        while(left <= right)
        {
            mid = (left + right) >> 1;
            if(check(mid,n))
            {
                left = mid + 1;
                ans = mid ;
            }
            else  right = mid - 1;
        }
        if(ans < 4) printf("%d\n",ans);
        else printf("%d\n",ans+1);
    }
    return 0;
}