poj1743之后缀数组

Musical Theme

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 15932		Accepted: 5509

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.

题目意思是求不重叠的最长相同变化的子串，输出该长度

比如1 2 3 4 5 6 7 8 9 10,最长长度为1，因为子串1 2 3 4 5 和 6 7 8 9 10变化都一样的

思路：既然要求变化一样，那么可以让原数组前后相减，然后利用后缀数组height的性质求子串最长公共前缀即可

height性质：1.height[i]表示排名为i和i-1的子串的最长公共前缀

2.由于最相似的子串排名肯定是最接近的，所以height的大小肯定是这样的：大-》小，大-》小，大-》小......

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=20000+10;
int *rank,r[MAX],sa[MAX],height[MAX];
int wa[MAX],wb[MAX],wm[MAX];

bool cmp(int *r,int a,int b,int l){
	return r[a] == r[b] && r[a+l] == r[b+l];
}

void makesa(int *r,int *sa,int n,int m){
	int *x=wa,*y=wb,*t;
	for(int i=0;i<m;++i)wm[i]=0;
	for(int i=0;i<n;++i)wm[x[i]=r[i]]++;
	for(int i=1;i<m;++i)wm[i]+=wm[i-1];
	for(int i=n-1;i>=0;--i)sa[--wm[x[i]]]=i;
	for(int i=0,j=1,p=0;p<n;j=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<m;++i)wm[i]=0;
		for(i=0;i<n;++i)wm[x[y[i]]]++;
		for(i=1;i<m;++i)wm[i]+=wm[i-1];
		for(i=n-1;i>=0;--i)sa[--wm[x[y[i]]]]=y[i];
		for(t=x,x=y,y=t,i=p=1,x[sa[0]]=0;i<n;++i){
			x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++;
		}
	}
	rank=x;
}

void calheight(int *r,int *sa,int n){
	for(int i=0,j=0,k=0;i<n;height[rank[i++]]=k){
		for(k?--k:0,j=sa[rank[i]-1];r[i+k] == r[j+k];++k);
	}
}

int main(){
	int n;
	while(cin>>n,n){
		for(int i=0;i<n;++i)cin>>r[i];
		for(int i=0;i<n-1;++i)r[i]=r[i+1]-r[i]+100; 
		r[--n]=0;
		makesa(r,sa,n+1,200);
		calheight(r,sa,n);//这里注意不要传递n+1过去,rank[n]-1=0-1=-1 
		int l=1,r=n+1,mid,L,R;
		bool flag=false;
		while(l<=r){
			mid=l+r>>1;
			L=INF,R=-INF,flag=false;
			for(int i=1;i<=n;++i){
				if(height[i]>=mid){
					L=min(L,sa[i]);
					L=min(L,sa[i-1]);
					R=max(R,sa[i]);
					R=max(R,sa[i-1]);
				}else{
					if(L+mid+1<=R)flag=true;//这里注意是L+mid+1,因为r[L+mid]=r[L+mid+1]-r[L+mid],从L开始原串中前面那个子串包括L+mid+1 
					L=INF,R=-INF;
				}
			}
			if(L+mid+1<=R)flag=true;
			if(flag)l=mid+1;
			else r=mid-1;
		}
		if(l>=5)cout<<l<<endl;
		else cout<<0<<endl;
	}
	return 0;
}