poj 1743 Musical Theme

Musical Theme

Time Limit: 1000MS		Memory Limit: 30000K

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

Hint

Use scanf instead of cin to reduce the read time.

Source

LouTiancheng@POJ

题目大意：

从所给字符串里找最长的2个子串，这2个子串满足以下3个条件：

① 每个长度>=5

② 2个子串的第i个数到第i+1个数的加减变化情况相等

③ 2子串不重叠

输出子串最长长度

多组数据

解题思路：

对于条件② ，差分，在差分序列上做后缀数组

求最值，可以二分答案mid，

满足条件①，二分起点l=4，因为差分数组长度为len，对应的序列长度为len+1

后缀数组得到的height数组，根据mid分组，保证相邻的 height值>=mid的分在同一组

这样分出的组满足：同一组内任意两后缀的公共前缀>=mid

每一组内，找到后缀起始位置的min，max，判断是否满足条件③

输出ans+1，原因同l=4

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define N 20010  
using namespace std;
int n,p,q=1,k,a[N],rk[2][N],sa[2][N],v[N],h[N],ans=-1,f[N][16];
char s[N]; 
void mul(int *sa,int *rk,int *SA,int *RK)
{
    for(int i=1;i<=n;i++) v[rk[sa[i]]]=i;
    for(int i=n;i;i--) if(sa[i]>k) SA[v[rk[sa[i]-k]]--]=sa[i]-k;
    for(int i=n-k+1;i<=n;i++) SA[v[rk[i]]--]=i;
    for(int i=1;i<=n;i++) RK[SA[i]]=RK[SA[i-1]]+(rk[SA[i]]!=rk[SA[i-1]]||rk[SA[i]+k]!=rk[SA[i-1]+k]); 
} 
void presa()
{
    for(int i=1;i<=n;i++) v[a[i]]++;
    for(int i=1;i<=180;i++) v[i]+=v[i-1];
    for(int i=1;i<=n;i++) sa[p][v[a[i]]--]=i;
    for(int i=1;i<=n;i++) rk[p][sa[p][i]]=rk[p][sa[p][i-1]]+(a[sa[p][i]]!=a[sa[p][i-1]]);
    for(k=1;k<n;k<<=1,swap(p,q)) 
    {
        mul(sa[p],rk[p],sa[q],rk[q]);
        if(rk[q][sa[q][n]]==n) 
        {
            swap(p,q);
            return;
        }
    }
}
void getheight()
{
    int j,g=0;
    for(int i=1;i<=n;i++)
    {
        j=sa[p][rk[p][i]-1];
        while(a[i+g]==a[j+g]) g++;
        h[rk[p][i]]=g;if(g) g--;
    }
}
void pre()
{
    p=0;q=1;k=0;
    memset(a,0,sizeof(a));
    memset(rk,0,sizeof(rk));
    memset(sa,0,sizeof(sa));
    memset(v,0,sizeof(v));
    memset(h,0,sizeof(h));
    ans=-1;
}
bool check(int mid)
{
    int minn=n+1,maxn=-1;
    for(int i=2;i<=n;i++)
    {
        if(h[i]<mid)
        {
            if(maxn-minn>=mid) return true;
            minn=n+1;maxn=-1;
        }
        minn=min(minn,sa[p][i]);
        maxn=max(maxn,sa[p][i]);
    }
    if(maxn-minn>=mid) return true;
    return false;
}
void solve()
{
    int l=4,r=n,mid;
    while(l<=r)
    {
        mid=l+r>>1;
        if(check(mid)) ans=mid,l=mid+1;
        else r=mid-1;
    }
    printf("%d
",ans+1);
}
int main()
{
    while(scanf("%d",&n))
    {
        if(!n) return 0;
        pre();
        int last,x;
        scanf("%d",&last);
        for(int i=2;i<=n;i++)
        {
            scanf("%d",&x);
            a[i-1]=x-last;
            a[i-1]+=88;
            last=x;
        }
        n--;
        presa();
        getheight();
        solve();
    }
}