poj 1743 后缀数组求最长不重叠重复子串
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Musical Theme
Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 19089 Accepted: 6545
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:
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!
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.
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:
思路:后缀数组。求出任意相邻音符的差值,然后把问题转化为 不可重叠最长重复子串,用后缀数组来做。先二分答案,把题目变成判定性问题:判断是否存在两个长度为k的子串是相同的,且不重叠。解决这个问题的关键还是利用 height数组。把排序后的后缀分成若干组,其中每组的后缀之间的height值都不小于k。例如,字符串为“aabaaaab”,当k=2时,后缀分成了4组,如下图所示(摘自罗穗骞的国家集训队论文):
ps:在每个分组里面找不重叠的子串
#include<stdio.h>#include<iostream>#define N 20005using namespace std;int t1[N],t2[N],x[N],s[N],c[N],sa[N],height[N],rank[N];void build_sa(int *s,int n,int m){ int *x=t1,*y=t2,i,k,p; for(i=0;i<m;i++) c[i]=0; for(i=0;i<n;i++) c[x[i]=s[i]]++; for(i=1;i<m;i++) c[i]+=c[i-1]; for(i=n-1;i>=0;i--) sa[--c[x[i]]]=i; for(k=1;k<=n;k<<=1) { p=0; for(i=n-k;i<n;i++) y[p++]=i; for(i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k; for(i=0;i<m;i++) c[i]=0; for(i=0;i<n;i++) c[x[y[i]]]++; for(i=1;i<m;i++) c[i]+=c[i-1]; for(i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i]; swap(x,y); p=1; x[sa[0]]=0; for(i=1;i<n;i++) x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++; if(p>=n) break; m=p; }}void getheight(int n){ int i,k=0,j; for(i=0;i<=n;i++) rank[sa[i]]=i; for(i=0;i<n;i++) { if(k) k--; j=sa[rank[i]-1]; while(s[i+k]==s[j+k]) k++; height[rank[i]]=k; }}int check(int k,int n){ int ma,mi,i; ma=mi=sa[1]; for(i=2;i<=n;i++) { if(height[i]<k) ma=mi=sa[i]; else { if(sa[i]>ma) ma=sa[i]; if(sa[i]<mi) mi=sa[i]; if(ma-mi>=k) return 1; } } return 0;}int main(){ int n,i,l,r,ans; while(scanf("%d",&n),n) { for(i=0;i<n;i++) scanf("%d",&s[i]); for(i=n-1;i>0;i--) s[i]=s[i]-s[i-1]+90; for(i=0;i<n-1;i++) s[i]=s[i+1]; n--; s[n]=0; build_sa(s,n+1,180); getheight(n); l=0;r=n/2; ans=0; while(l<=r) { int mid=(l+r)>>1; if(check(mid,n)==1) { ans=mid; l=mid+1; } else r=mid-1; } if(ans<4) ans=0; else ans++; printf("%d\n",ans); } return 0;}
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