HDU2559-Longest Repeated subsequence(后缀数组)

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本文介绍了一种寻找序列中至少重复K次的最长子序列的算法，并提供了完整的C++实现代码。该算法首先通过排序和唯一化处理输入序列，然后使用后缀数组和区间最小查询等数据结构来高效解决问题。

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Longest Repeated subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 951 Accepted Submission(s): 188

Problem Description

Write a program that takes a sequence of number and returns length of the longest repeated
subsequence. A repeated subsequence which repeats identically at least K times without overlaps.
For example 1 2 3 1 2 3 2 3 1 repeats 2 3 for three times.

Input

Line 1: The input contains a single integer T , the number of test cases.
Line 2: Two space-separated integers: N and K (1 <= n <= 50000), (2 <= k <= n)
Lines 3..N+2: N integers, one per line. The integer is between 0 and 10^9.
It is guaranteed that at least one subsequence is repeated at least K times.

Output

For each test case, the first line print the length n of the Longest Repeated Subsequence and the line 2…n+1 print the subsequence numbers.(if we have several subsequence with the same length output the lexicographic order smallest one). Separate output for adjacent cases with a single blank line.

Sample Input

Sample Output

2
2
3

题意:给一个含n个数的序列，求最少重复k次的最长子序列（子序列不能重叠）。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>

using namespace std;

const int N=200010;
const int INF=0x7FFFFFFF;

struct Sa
{
    int s[N],a[N],b[N];
    int rk[2][N],sa[N],h[N],w[N],now,k,m,n;
    int rmq[N][20],lg[N],bel[N];

    bool GetS()
    {
        scanf("%d %d",&n,&k);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]),s[i]=a[i];
        sort(a+1,a+1+n);
        m=unique(a+1,a+1+n)-a;
        for(int i=1;i<=n;i++)
            s[i]=lower_bound(a+1,a+m,s[i])-a;
        return true;
    }

    void getsa(int z,int &m)
    {
        int x=now,y=now^=1;
        for(int i=1; i<=z; i++) rk[y][i]=n-i+1;
        for(int i=1,j=z; i<=n; i++)
            if(sa[i]>z) rk[y][++j]=sa[i]-z;
        for(int i=1; i<=m; i++) w[i]=0;
        for(int i=1; i<=n; i++) w[rk[x][rk[y][i]]]++;
        for(int i=1; i<=m; i++) w[i]+=w[i-1];
        for(int i=n; i>=1; i--) sa[w[rk[x][rk[y][i]]]--]=rk[y][i];
        for(int i=m=1; i<=n; i++)
        {
            int *a=rk[x]+sa[i],*b=rk[x]+sa[i-1];
            rk[y][sa[i]]=*a==*b&&*(a+z)==*(b+z)?m-1:m++;
        }
    }

    void getsa(int m)
    {
        now=rk[1][0]=sa[0]=s[0]=0;
        for(int i=1; i<=m; i++) w[i]=0;
        for(int i=1; i<=n; i++) w[s[i]]++;
        for(int i=1; i<=m; i++) rk[1][i]=rk[1][i-1]+(bool)w[i];
        for(int i=1; i<=m; i++) w[i]+=w[i-1];
        for(int i=1; i<=n; i++) rk[0][i]=rk[1][s[i]];
        for(int i=1; i<=n; i++) sa[w[s[i]]--]=i;
        rk[1][n+1]=rk[0][n+1]=0;
        for(int x=1,y=rk[1][m]; x<=n&&y<=n; x<<=1) getsa(x,y);
        for(int i=1,j=0; i<=n; h[rk[now][i++]]=j?j--:j)
        {
            if(rk[now][i]==1) continue;
            int k=n-max(sa[rk[now][i]-1],i);
            while(j<=k&&s[sa[rk[now][i]-1]+j]==s[i+j]) ++j;
        }
    }

    void getrmq()
    {
        h[n+1]=h[1]=lg[1]=0;
        for(int i=2; i<=n; i++)
            rmq[i][0]=h[i],lg[i]=lg[i>>1]+1;
        for(int i=1; (1<<i)<=n; i++)
        {
            for(int j=2; j<=n; j++)
            {
                if(j+(1<<i)>n+1) break;
                rmq[j][i]=min(rmq[j][i-1],rmq[j+(1<<i-1)][i-1]);
            }
        }
    }

    int lcp(int x,int y)
    {
        int l=min(rk[now][x],rk[now][y])+1,r=max(rk[now][x],rk[now][y]);
        return min(rmq[l][lg[r-l+1]],rmq[r-(1<<lg[r-l+1])+1][lg[r-l+1]]);
    }

    int check(int x,int &y)
    {
    	int sum=0;
    	b[sum++]=sa[1];
    	for(int i=2;i<=n+1;i++)
        {
            if(i<=n&&h[i]>=x) b[sum++]=sa[i];
            else
            {
                sort(b,b+sum);
                int t=b[0],cnt=1;
                for(int j=1;j<sum;j++)
                    if(b[j]>=x+t) t=b[j],cnt++;
                if(cnt>=k) {y=sa[i-1];return true;}
                sum=0;b[sum++]=sa[i];
            }
        }
        return false;
	}

    void work()
    {
        getsa(m);
        int l=1,r=n,g;
        while(l<=r)
        {
            if(check(l+r>>1,g)) l=(l+r>>1)+1;
            else r=(l+r>>1)-1;
        }
        check(r,g);
        printf("%d\n",r);
        for(int i=1;i<=r;i++)
            printf("%d\n",a[s[g+i-1]]);
    }
}sa;

int main()
{
    int t,cas=0;
    scanf("%d",&t);
    while(t--)
    {
        if(cas) printf("\n");
        cas=1;
        sa.GetS();
        sa.work();
    }
    return 0;
}