POJ 1743 后缀数组

 

Musical Theme

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 24507		Accepted: 8259

 

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.

/*
POJ 1743  后缀数组

给你一串数字，求它们最长的重复(公差相同)子序列，且两个子序列不相交

表示第一次用后缀数组
我们可以得到
sa[i]:表示排名第i个的首字母位置
Rank[i]:第i个数的排名
Height[i]:sa[i]和sa[i-1]的最长公共前缀


于是我们先求出相邻两个数之间的差，然后二分来枚举长度
如果在当前长度下能找到两个并且相距合适的就记录下来
因为最后得到的是间距的个数  所以ans+1

*/
#include <algorithm>
#include <cmath>
#include <queue>
#include <iostream>
#include <cstring>
#include <map>
#include <cstdio>
#include <vector>
#include <functional>
#define lson (i<<1)
#define rson ((i<<1)|1)
using namespace std;
typedef long long ll;
const int maxn = 20050;

int t1[maxn],t2[maxn],c[maxn];
bool cmp(int *r,int a,int b,int l)
{
    return r[a]==r[b] &&r[l+a] == r[l+b];
}

void get_sa(int str[],int sa[],int Rank[],int height[],int n,int m)
{
    n++;
    int p,*x=t1,*y=t2;
    for(int i = 0; i < m; i++) c[i] = 0;
    for(int i = 0; i < n; i++) c[x[i] = str[i]]++;
    for(int i = 1; i < m; i++) c[i] += c[i-1];
    for(int i = n-1; i>=0; i--) sa[--c[x[i]]] = i;
    for(int j = 1; j <= n; j <<= 1)
    {
        p = 0;
        for(int i = n-j; i < n; i++) y[p++] = i;
        for(int i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;
        for(int i = 0; i < m; i++) c[i] = 0;
        for(int i = 0; i < n; i++) c[x[y[i]]]++ ;
        for(int i = 1; i < m; i++) c[i] += c[i-1];
        for(int i = n-1; i >= 0; i--)  sa[--c[x[y[i]]]] = y[i];

        swap(x,y);
        p = 1;
        x[sa[0]] = 0;
        for(int 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(int i = 0; i <= n; i++)
        Rank[sa[i]] = i;
    for(int i = 0; i < n; i++)
    {
        if(k) k--;
        int j = sa[Rank[i]-1];
        while(str[i+k] == str[j+k]) k++;
        height[Rank[i]] = k;
    }

}

int Rank[maxn],height[maxn];
int sa[maxn];
int a[maxn];

bool judge(int len,int n)
{
    for(int i = 2; i <= n; i++)
    {
        if(height[i] < len)
            continue;
        for(int j = i-1; j >= 2; j--)
        {
            if(abs(sa[i] - sa[j]) >= len)
                return true;
            if(height[j] < len)
                break;
        }
    }
    return false;
}

int main()
{
    int n;
    while(scanf("%d",&n) != EOF)
    {

        if(!n)
            break;
        for(int i = 0; i < n; i++)  scanf("%d",&a[i]);
        for(int i = 0; i < n; i++)
            a[i] = a[i+1]-a[i]+90;
        a[n-1] = 0;
        if(n < 10)
        {
            printf("0\n");
            continue;
        }
        get_sa(a,sa,Rank,height,n-1,200);

        int l = 0,r = n;
        int ans ;
        while(l <= r)
        {
            int mid = (l+r) >>1;
            if(judge(mid,n))
            {
                ans = mid;
                l = mid+1;
            }
            else
                r = mid-1;
        }
        if(ans < 4)
            printf("0\n");
        else
            printf("%d\n",ans+1);
    }
    return 0;
}