poj 1743 Musical Theme(最长不重叠重复子串 后缀数组+二分)
题目：http://poj.org/problem?id=1743
题意：给你一个序列，求序列里长度至少为5的最长不重叠重复子串，这里的重复包括每个元素加减相同的值
分析：对于区间加减可以通过将相邻的元素的差组成新的序列，答案就变成求至少长度4的最长不重叠重复子串，并且至少间隔1的距离，然后就可以用后缀数组来做了，然后二分子串的长度L，找到height数组大于等于这个长度L的区间，求区间对应原序列的下标的最大值high和最小值low，只要high-low>L就表示存在长度为L的解。。。
PS：这题居然是楼教主的男人八题之一。。。当初就应该把他们都A了= =
代码：
/** head files*/
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;
/** some operate*/
#define PB push_back
#define MP make_pair
#define REP(i,n) for(i=0;i<(n);++i)
#define UPTO(i,l,h) for(i=(l);i<=(h);++i)
#define DOWN(i,h,l) for(i=(h);i>=(l);--i)
#define MSET(arr,val) memset(arr,val,sizeof(arr))
#define MAX3(a,b,c) max(a,max(b,c))
#define MAX4(a,b,c,d) max(max(a,b),max(c,d))
#define MIN3(a,b,c) min(a,min(b,c))
#define MIN4(a,b,c,d) min(min(a,b),min(c,d))
/** some const*/
#define N 22222
#define M 222222
#define PI acos(-1.0)
#define oo 1111111111
/** some alias*/
typedef long long ll;
/** Global variables*/
/** some template names, just push ctrl+j to get it in*/
//manacher 求最长回文子串
//pqueue 优先队列
//combk n元素序列的第m小的组合和
//pmatrix n个点的最大子矩阵
//suffixarray 后缀数组
template <typename T, int LEN>
struct suffixarray
{
int str[LEN*3],sa[LEN*3];
int rank[LEN],height[LEN];
int id[LEN];
int len;
bool equal(int *str, int a, int b)
{
return str[a]==str[b]&&str[a+1]==str[b+1]&&str[a+2]==str[b+2];
}
bool cmp3(int *str, int *nstr, int a, int b)
{
if(str[a]!=str[b])return str[a]<str[b];
if(str[a+1]!=str[b+1])return str[a+1]<str[b+1];
return nstr[a+b%3]<nstr[b+b%3];
}
void radixsort(int *str, int *sa, int *res, int n, int m)
{
int i;
REP(i,m)id[i]=0;
REP(i,n)++id[str[sa[i]]];
REP(i,m)id[i+1]+=id[i];
DOWN(i,n-1,0)res[--id[str[sa[i]]]]=sa[i];
}
void dc3(int *str, int *sa, int n, int m)
{
#define F(x) ((x)/3+((x)%3==1?0:one))
#define G(x) ((x)<one?(x)*3+1:((x)-one)*3+2)
int *nstr=str+n, *nsa=sa+n, *tmpa=rank, *tmpb=height;
int i,j,k,len=0,num=0,zero=0,one=(n+1)/3;
REP(i,n)if(i%3)tmpa[len++]=i;
str[n]=str[n+1]=0;
radixsort(str+2, tmpa, tmpb, len, m);
radixsort(str+1, tmpb, tmpa, len, m);
radixsort(str+0, tmpa, tmpb, len, m);
nstr[F(tmpb[0])]=num++;
UPTO(i,1,len-1)
nstr[F(tmpb[i])]=equal(str,tmpb[i-1],tmpb[i])?num-1:num++;
if(num<len)dc3(nstr,nsa,len,num);
else REP(i,len)nsa[nstr[i]]=i;
if(n%3==1)tmpa[zero++]=n-1;
REP(i,len)if(nsa[i]<one)tmpa[zero++]=nsa[i]*3;
radixsort(str, tmpa, tmpb, zero, m);
REP(i,len)tmpa[nsa[i]=G(nsa[i])]=i;
i=j=0;
REP(k,n)
if(j>=len||(i<zero&&cmp3(str,tmpa,tmpb[i],nsa[j])))sa[k]=tmpb[i++];
else sa[k]=nsa[j++];
}
void initSA(T *s, int n,int m)
{
int i,j,k=0;
str[len=n]=0;
REP(i,n)str[i]=s[i];
dc3(str,sa,n+1,m);
REP(i,n)sa[i]=sa[i+1];
REP(i,n)rank[sa[i]]=i;
REP(i,n)
{
if(k)--k;
if(rank[i])for(j=sa[rank[i]-1];str[i+k]==str[j+k];++k);
else k=0;
height[rank[i]]=k;
}
}
};
suffixarray<int,N> msa;
int s[N];
bool ok(int m)
{
int i=0,j,low,high;
while(i<msa.len)
{
while(i<msa.len&&msa.height[i]<m)++i;
j=i;
low=msa.len,high=-1;
low=min(low,msa.sa[i-1]);
high=max(high,msa.sa[i-1]);
while(j<msa.len&&msa.height[j]>=m)
{
low=min(low,msa.sa[j]);
high=max(high,msa.sa[j]);
++j;
}
if(high-low>m)return 1;
i=j;
}
return 0;
}
int main()
{
int i,n,ans;
while(scanf("%d",&n),n!=0)
{
REP(i,n)scanf("%d",&s[i]);
REP(i,n)s[i]=200+s[i]-s[i+1];
msa.initSA(s,n-1,555);
int l=4,r=n,m;
ans=0;
while(l<=r)
{
m=(l+r)>>1;
if(ok(m))ans=l=m+1;
else r=m-1;
}
printf("%d\n",ans);
}
return 0;
}