poj_1743_Musical Theme(后缀数组)

题目链接：poj_1743_Musical Theme

题意：

给你一串数字，让你找最长的变化相同不重叠的子串，至少长度为5

题解：

处理数据后用后缀数组加二分答案，然后用height数组check答案，运用height数组求相同不重叠的子串经典运用

#include<cstdio>
#include<algorithm>
#define F(i,a,b) for(int i=a;i<=b;i++)
using namespace std;

namespace suffixarray{    
    #define FN(n) for(int i=0;i<n;i++)
    const int N =2E4+7;
    int rnk[N],sa[N],height[N],c[N],s[N];
    void getsa(int n,int m,int *x=rnk,int *y=height){
        FN(m)c[i]=0;FN(n)c[x[i]=s[i]]++;FN(m)c[i+1]+=c[i];
        for(int i=n-1;i>=0;i--)sa[--c[x[i]]]=i;
        for(int k=1,p;p=0,k<=n;k=p>=n?N:k<<1,m=p){
            for(int i=n-k;i<n;i++)y[p++]=i;
            FN(n)if(sa[i]>=k)y[p++]=sa[i]-k;
            FN(m)c[i]=0;FN(n)c[x[y[i]]]++;FN(m)c[i+1]+=c[i];
            for(int i=n-1;i>=0;i--)sa[--c[x[y[i]]]]=y[i];
            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]+k]==y[sa[i]+k]?p-1:p++;
        }
        FN(n)rnk[sa[i]]=i;
        for(int i=0,j,k=0;i<n-1;height[rnk[i++]]=k)
        for(k=k?k-1:k,j=sa[rnk[i]-1];s[i+k]==s[j+k];k++);
    }
}

using namespace suffixarray;
int n;
inline void upd(int &a,int b){if(a>b)a=b;}
inline void upu(int &a,int b){if(a<b)a=b;}

inline bool check(int x)
{
    int l=N,r=0;
    F(i,1,n)
    {
        if(height[i]>=x)
        {
            upd(l,sa[i]),upu(r,sa[i]);
            if(r-l>=x)return 1;
        }else l=r=sa[i];
    }
    return 0;
}

int main()
{
    while(scanf("%d",&n),n)
    {
        F(i,0,n-1)scanf("%d",s+i);
        if(n<10){puts("0");continue;}
        F(i,0,n-2)s[i]=s[i+1]-s[i]+89;
        s[--n]=0,getsa(n+1,200);
        int l=4,r=n,mid;
        while(l<=r)mid=(l+r)>>1,check(mid)?l=mid+1:r=mid-1;
        printf("%d
",l<5?0:l);
    }
    return 0;
}