POJ1743之后缀数组

Musical Theme
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 25579 Accepted: 8630
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.

给出一些音符，求出最长的重复出现的旋转长度。

只要满足相邻的差相等便可以了。

那我们建立一个相邻并非的数组，题目要求的便是求最长的重复子串长度，而且不可重叠。

由于相邻差可能为负，则统一加上100，转变为0-200之间的数即可。

之后求出height数组，表示的是相邻后缀的最长公共前缀。

二分答案，然后进行判定。

按height就可以对后缀进行分组。
height如果大于长度，则我们需要判断是否重叠，通过sa的差值即可。

用了又一个模板>.<

#include <cstdio>#include <algorithm>#include <cstring>#include <iostream>using namespace std;const int N = int(2e5)+10;int a[N],b[N],sa[N];int cmp(int *r,int a,int b,int l){    return (r[a]==r[b]) && (r[a+l]==r[b+l]);}int wa[N],wb[N],WS[N],wv[N];int RANK[N],height[N];void DA(int *r,int *sa,int n,int m){    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]]++;    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++;        //printf("p = %d\n", p );    }}void calheight(int *r,int *sa,int n){    //  memset(height,0,sizeof(height));    //  memset(RANK,0,sizeof(RANK));    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++);}bool check(int m,int n){    int maxx=0,minn=0;    for(int i=1;i<=n;i++)    {        if(height[i]< m)        {            maxx=sa[i];            minn=sa[i];        }        else        {            maxx=max(maxx,max(sa[i],sa[i-1]));            minn=min(minn,min(sa[i],sa[i-1]));            if(maxx-minn>=m)//后缀不重叠                return true;        }    }    return false;}long long nxt[N];int main (void){    int t;   // cin>>t;    while(~scanf("%d",&t))    {        if(t==0)            break;        for(int i=0;i<t;i++)        {            scanf("%d",&a[i]);        }        for(int i=0;i<t-1;i++)        {            a[i]=a[i+1]-a[i]+100;        }        t--;        a[t]=0;        //get_sa(b,t);        //get_height(b,t);        DA(a,sa,t+1,200);        calheight(a,sa,t);        if(!check(4,t))        {            printf("0\n");            continue;        }        long long ans=0;        int l=0,r=t,mid;        while(r-l>1)        {            mid=(l+r)/2;            if(check(mid,t))            {                ans=mid;                l=mid;            }            else                r=mid;        }        printf("%lld\n",ans+1);    }    return 0;}