POJ 1743Musical Theme（后缀数组）

Musical Theme

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 30336 Accepted: 10156

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

3025 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 1882 78 74 70 66 67 64 60 65 800

Sample Output

5

Hint

Use scanf instead of cin to reduce the read time.

Source

LouTiancheng@POJ

题意：有N(1 <= N <=20000)个音符的序列来表示一首乐曲，每个音符都是1..88范围内的整数，现在要找一个重复的主题。“主题”是整个音符序列的一个子串，它需要满足如下条件：

    1.长度至少为5个音符。

    2.在乐曲中重复出现。(可能经过转调，“转调”的意思是主题序列中每个音符都被加上或减去了同一个整数值)

    3.重复出现的同一主题不能有公共部分。

想法：那么我们通过进行二分来求乐曲里的k值，求出height[i]大于k的时候，然后求出这个区间里面起始位置最大的差值，要求差值也要大于k，这样才保证没有重叠，这是因为height里的都是已经按照字典序排好的了

代码：

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define N 20005int wa[N],wb[N],wsf[N],wv[N],sa[N];int rank[N],height[N],s[N],a[N],n;char str1[N],str2[N];//sa:字典序中排第i位的起始位置在str中第sa[i]//rank:就是str第i个位置的后缀是在字典序排第几//height：字典序排i和i-1的后缀的最长公共前缀int cmp(int *r,int a,int b,int k){    return r[a]==r[b]&&r[a+k]==r[b+k];}void getsa(int *r,int *sa,int n,int m)//n要包含末尾添加的0{    int i,j,p,*x=wa,*y=wb,*t;    for(i=0; i<m; i++)  wsf[i]=0;    for(i=0; i<n; i++)  wsf[x[i]=r[i]]++;    for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];    for(i=n-1; i>=0; i--)  sa[--wsf[x[i]]]=i;    p=1;    j=1;    for(; 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++)  wsf[i]=0;        for(i=0; i<n; i++)  wsf[wv[i]]++;        for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];        for(i=n-1; i>=0; i--)  sa[--wsf[wv[i]]]=y[i];        t=x;        x=y;        y=t;        x[sa[0]]=0;        for(p=1,i=1; i<n; i++)            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;    }}void getheight(int *r,int n)//n不保存最后的0{    int i,j,k=0;    for(i=1; i<=n; i++)  rank[sa[i]]=i;    for(i=0; i<n; i++)    {        if(k)            k--;        else            k=0;        j=sa[rank[i]-1];        while(r[i+k]==r[j+k])            k++;        height[rank[i]]=k;    }}int ans;int fun(int k){    int i,maxn,minn;    maxn = minn = sa[1];    for(i=2;i<=n-1;i++)    {        if(height[i]>=k && i<n-1)        {            minn = min(minn,sa[i]);            maxn = max(maxn,sa[i]);            continue;        }        if(maxn-minn>=k) return 1;        maxn = minn = sa[i];    }    return 0;}int main(){    int i,j,k;    while((~scanf("%d",&n),n))    {        for(i=0;i<=n-1;i++)        {            scanf("%d",&s[i]);        }        for(i=0;i<=n-2;i++)        {            s[i] = s[i+1]-s[i]+100;        }        s[n-1] = 0;        getsa(s,sa,n,200);        getheight(s,n-1);        int l = 4,r = n-1;        while(l<=r)        {            int mid = (l+r)/2;            if(fun(mid))            {                ans = mid;                l=mid+1;            }            else r = mid-1;        }        ans++;        printf("%d\n",ans<5?0:ans);    }    return 0;}