poj 1743 Musical Theme

最新推荐文章于 2019-03-16 17:03:00 发布

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本文介绍了一种用于查找音乐旋律中最长重复主题的算法。通过将旋律转换为整数序列并利用后缀数组技术，该算法能够高效地找到满足条件的主题。文章提供了详细的代码实现和示例。

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Musical Theme

Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 28794 Accepted: 9708

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.

Source

LouTiancheng@POJ

【分析】
把题面转换一下，变为求最长不重叠重复子串…
思路不难想（但我也想不出来哈哈）…大家参考论文《后缀数组——处理字符串的有力工具》吧。

论文传送门

【代码】

//poj 1743
#include<iostream>
#include<cstdio>
#include<cstring>
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mxn=100005;
int n,m,len;
int sa[mxn],x[mxn],y[mxn],b[mxn],a[mxn],s[mxn],rank[mxn],height[mxn];
inline bool comp(int i,int j,int l) //比较rank是否相同 
{
    return y[i]==y[j]&&(i+l>len?-1:y[i+l])==(j+l>len?-1:y[j+l]);
}
inline void work()
{
    int i,j,k,p;
    fo(i,0,m) b[i]=0;  //把桶清零
    fo(i,1,len) b[x[i]=a[i]]++;
    fo(i,1,m) b[i]+=b[i-1];
    for(i=len;i>=1;i--) sa[b[x[i]]--]=i;  //第一次排序
    for(k=1;k<=len;k<<=1)
    {
        p=0;
        for(i=len-k+1;i<=len;i++) y[++p]=i;  //第二关键字为0的数 
        fo(i,1,len) if(sa[i]>k) y[++p]=sa[i]-k;  //与上一行构成第二关键字排序 
        fo(i,0,m) b[i]=0;
        fo(i,1,len) b[x[y[i]]]++;
        fo(i,1,m) b[i]+=b[i-1];
        for(i=len;i>=1;i--) sa[b[x[y[i]]]--]=y[i];
        swap(x,y);p=2;x[sa[1]]=1;  //y储存旧的rank
        fo(i,2,len)
          x[sa[i]]=comp(sa[i-1],sa[i],k)?p-1:p++;  //x储存新的rank 
        if(p>len) break;
        m=p-1; 
    }
    p=k=0;   
    fo(i,1,len) rank[sa[i]]=i;
    for(i=1;i<=len;height[rank[i++]]=k)
      for(k?k--:0,j=sa[rank[i]-1];a[i+k]==a[j+k];k++);
}
inline bool ok(int k)
{
    int i,j,mn=1e9,mx=-1e9;
    fo(i,2,len)
    {
        if(height[i]>=k)
        {
            mn=min(mn,min(sa[i],sa[i-1]));
            mx=max(mx,max(sa[i],sa[i-1]));
            if(mx-mn>k) return 1;
        }
        else mn=1e9,mx=-1e9;
    }
    return 0;
}
int main()
{
    int i,j,k;
    while(scanf("%d",&len) && len)
    {
        M(sa),M(x),M(y),M(rank),M(height),M(s),M(a),M(b);m=200;
        fo(i,1,len) scanf("%d",&s[i]);
        fo(i,1,len) a[i]=s[i]-s[i-1]+88;
        work();
        int l=0,r=len>>1;
        while(l<r)
        {
            int mid=l+r+1>>1;
            if(ok(mid)) l=mid;
            else r=mid-1;
        }
        if(l>=4) printf("%d\n",l+1);
        else printf("0\n");
    }
    return 0;
}
//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