Musical Theme
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 26997   Accepted: 9096

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

【解题方法】论文上的题目，也是楼教主的男人8题之一了。二分+后缀数组。

首先题中说了可以转调，那么我们取相邻位置的数值作为新数组进行运算，然后容易知道，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。然后对于每一组后缀数组，只需要判断每个后缀的sa值的最大值和最小值的差是否不小于K。如果有一组满足，则说明存在，否则不存在。

【复杂度分析】O(n*log(n))

【代码君】

//
//Created by just_sort 2016/10/16
//Copyright (c) 2016 just_sort.All Rights Reserved
//

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 22222;
int sa[maxn];//SA数组，表示将S的n个后缀从小到大排序后把排好序的
             //的后缀的开头位置顺次放入SA中
int t1[maxn],t2[maxn],c[maxn];//求SA数组需要的中间变量，不需要赋值
int Rank[maxn],height[maxn];
//待排序的字符串放在s数组中，从s[0]到s[n-1],长度为n,且最大值小于m,
//除s[n-1]外的所有s[i]都大于0，r[n-1]=0
//函数结束以后结果放在sa数组中
void build_sa(int s[],int n,int m)
{
    int i,j,p,*x=t1,*y=t2;
    //第一轮基数排序，如果s的最大值很大，可改为快速排序
    for(i=0;i<m;i++)c[i]=0;
    for(i=0;i<n;i++)c[x[i]=s[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;
        //直接利用sa数组排序第二关键字
        for(i=n-j;i<n;i++)y[p++]=i;//后面的j个数第二关键字为空的最小
        for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;
        //这样数组y保存的就是按照第二关键字排序的结果
        //基数排序第一关键字
        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];
        //根据sa和x数组计算新的x数组
        swap(x,y);
        p=1;x[sa[0]]=0;
        for(i=1;i<n;i++)
            x[sa[i]]=y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;
        if(p>=n)break;
        m=p;//下次基数排序的最大值
    }
}
void getHeight(int s[],int n)
{
    int i,j,k=0;
    for(i=0;i<=n;i++)Rank[sa[i]]=i;
    for(i=0;i<n;i++)
    {
        if(k)k--;
        j=sa[Rank[i]-1];
        while(s[i+k]==s[j+k])k++;
        height[Rank[i]]=k;
    }
}

int s[maxn];
bool isok(int n, int t)
{
    int left = sa[1], right = sa[1];
    for(int i = 2; i <= n; i++)
    {
        if(height[i] < t) left = right = sa[i];
        else{
            left = min(left, sa[i]);
            right = max(right, sa[i]);
            if(right - left > t) return true;
        }
    }
    return false;
}
int main()
{
    int n;
    while(scanf("%d", &n) && n)
    {
        for(int i = 0; i < n; i++) scanf("%d", &s[i]);
        for(int i = n - 1; i > 0; i--) s[i] = s[i] - s[i-1] + 90;
        n--;
        for(int i = 0; i < n; i++) s[i] = s[i+1];
        s[n] = 0;
        build_sa(s, n + 1, 200);
        getHeight(s, n);
        int l = 1, r = n/2;
        int ans = -1;
        while(l <= r)
        {
            int mid = (l + r)/2;
            if(isok(n, mid)){
                ans = mid;
                l = mid + 1;
            }
            else{
                r = mid - 1;
            }
        }
        if(ans < 4){
            printf("0\n");
        }
        else{
            printf("%d\n", ans + 1);
        }
    }
    return 0;
}