Musical Theme
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 26374 | Accepted: 8902 |
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.
这道题是不能重叠的，于是二分check时不能只看height数组，把符合条件的后缀的的最大最小值保存下来，如果大于等于len，说明存在，否则不存在
/*
* POJ 1743 Musical Theme
* 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。
*
* 先转化成相邻两项的差值，就将转调的子串变成一样的了，然后就是找不可重叠重复子串。
* 做法就是二分答案LEN
* 然后根据height值进行分组
*/
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN = 20000+100;
/*
*suffix array
*倍增算法 O(n*logn)
*待排序数组长度为n,放在0~n-1中，在最后面补一个0
*build_sa( ,n+1, );//注意是n+1;
*getHeight(,n);
*例如：
*n = 8;
*num[] = { 1, 1, 2, 1, 1, 1, 1, 2, $ };注意num最后一位为0，其他大于0
*rank[] = { 4, 6, 8, 1, 2, 3, 5, 7, 0 };rank[0~n-1]为有效值，rank[n]必定为0无效值
*sa[] = { 8, 3, 4, 5, 0, 6, 1, 7, 2 };sa[1~n]为有效值，sa[0]必定为n是无效值
*height[]= { 0, 0, 3, 2, 3, 1, 2, 0, 1 };height[2~n]为有效值
*
*/
int sa[MAXN];//后缀数组，表示将S的n个后缀从小到大排序后把排好序的
//的后缀的开头位置顺次放入SA中
int t1[MAXN],t2[MAXN],c[MAXN];//求SA数组需要的中间变量，不需要赋值
int rank[MAXN],height[MAXN];
//待排序的字符串放在s数组中，从s[0]到s[n-1],长度为n,且最大值小于m,
//除s[n-1]外的所有s[i]都大于0，r[n-1]=0
//函数结束以后结果放在sa数组中
void build_sa(int s[],int n,int m)
{
int p,*x=t1,*y=t2;
//第一轮基数排序，如果s的最大值很大，可改为快速排序
for(int i=0;i<m;i++) c[i]=0;
for(int i=0;i<n;i++) c[x[i]=s[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)
{
int p=0;
//直接利用sa数组排序第二关键字
for(int i=n-j;i<n;i++) y[p++]=i;//后面的j个数第二关键字为空的最小
for(int i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;
//这样数组y保存的就是按照第二关键字排序的结果
//基数排序第一关键字
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];
//根据sa和x数组计算新的x数组
swap(x,y);
p=1;x[sa[0]]=0;
for(int i=1;i<n;i++)
x[sa[i]]=y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;
if(p>=n)break;
m=p;//下次基数排序的最大值
}
}
void getHeight(int s[],int n)
{
int k=0;
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(s[i+k]==s[j+k]) k++;
height[rank[i]]=k;
}
}
int s[MAXN];
bool check(int n,int len)
{
int ma=sa[1],mi=sa[1];
for(int i=2;i<=n;i++)
{
if(height[i]<len) ma=mi=sa[i];
else
{
mi=min(mi,sa[i]);
ma=max(ma,sa[i]);
if(ma-mi>len) return true;
}
}
return false;
}
int main()
{
int n;
while(scanf("%d",&n)==1&&n)
{
for(int i=0;i<n;i++) scanf("%d",&s[i]);
for(int i=n-1;i>0;i--) s[i]=s[i]-s[i-1]+90;
n--;
for(int i=0;i<n;i++) s[i]=s[i+1];
s[n]=0;
build_sa(s,n+1,200);
getHeight(s,n);
int ans=-1;
int l=1,r=n/2;
while(l<=r)
{
int mid=(l+r)/2;
if(check(n,mid)) ans=mid,l=mid+1;
else r=mid-1;
}
if(ans<4) printf("0\n");
else printf("%d\n",ans+1);
}
return 0;
}