HDU-5324 cdq分治

Boring Class

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1444    Accepted Submission(s): 460

Problem Description

Mr. Zstu and Mr. Hdu are taking a boring class , Mr. Zstu comes up with a problem to kill time, Mr. Hdu thinks it’s too easy, he solved it very quickly, what about you guys?
Here is the problem:
Give you two sequences L1,L2,...,Ln and R1,R2,...,Rn.
Your task is to find a longest subsequence v1,v2,...vm satisfies
v1≥1,vm≤n,vi<vi+1 .(for i from 1 to m - 1)
Lvi≥Lvi+1,Rvi≤Rvi+1(for i from 1 to m - 1)
If there are many longest subsequence satisfy the condition, output the sequence which has the smallest lexicographic order.

 

Input

There are several test cases, each test case begins with an integer n.
1≤n≤50000
Both of the following two lines contain n integers describe the two sequences.
1≤Li,Ri≤109

 

Output

For each test case ,output the an integer m indicates the length of the longest subsequence as described.
Output m integers in the next line.

 

Sample Input

55 4 3 2 16 7 8 9 1021 23 4

 

Sample Output

51 2 3 4 511

 

Author

ZSTU

 

Source

2015 Multi-University Training Contest 3

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5324

题目大意：

给出两个长度相同数列，求第一个不上升，第二个不下降的最长子序列长度。

解题思路：

二维的偏序关系问题，典型的cdq分治，稍微注意输出字典序最小的序列即可。

AC代码：

import java.util.*;public class Main {static int n,ct,ans,pos,x;static int[] aa=new int[50005];static int[] bb=new int[50005];static int[] cc=new int[50005];static int[] bit=new int[50005];static Pair[] pp=new Pair[50005];static Map<Integer,Integer> mp=new HashMap<Integer,Integer>();static void updata(int i,int x){for(;i<=ct;i+=i&-i)bit[i]=Math.max(bit[i],x);}static int max(int i){int res=0;for(;i>0;i-=i&-i)res=Math.max(res,bit[i]);return res;}static void clear(int i){for(;i<=ct;i+=i&-i)bit[i]=0;}static class Pair implements Comparable<Pair>{int x,y,z;Pair(int a,int b,int c){x=a;y=b;z=c;}public int compareTo(Pair p) {if(y==p.y) return z-p.z;return y-p.y;}}static void cdq(int l,int r){if(l==r) { cc[l]++;return;}int mid=(l+r)/2;cdq(mid+1,r);for(int i=l;i<=r;i++)pp[i]=new Pair(aa[i],bb[i],i);Arrays.sort(pp,l,r+1);;for(int i=r;i>=l;i--){x=pp[i].x;pos=pp[i].z;if(pos<=mid)cc[pos]=Math.max(cc[pos],max(x));else updata(x,cc[pos]);}for(int i=l;i<=r;i++)clear(pp[i].x);cdq(l,mid);}public static void main(String[] args) {Scanner in=new Scanner(System.in);while(in.hasNext()){n=in.nextInt();for(int i=1;i<=n;i++)aa[i]=cc[i]=in.nextInt();for(int i=1;i<=n;i++)bb[i]=in.nextInt();ct=0;Arrays.sort(cc,1,n+1);mp.clear();for(int i=1;i<=n;i++)if(cc[i]!=cc[i-1]) mp.put(cc[i],++ct);for(int i=1;i<=n;i++)aa[i]=mp.get(aa[i]);Arrays.fill(cc,0);Arrays.fill(bit,0);cdq(1,n);ans=pos=0;for(int i=1;i<=n;i++)if(cc[i]>ans) { ans=cc[i];pos=i;}System.out.println(ans);System.out.print(pos);for(int i=pos+1;i<=n;i++){if(cc[i]==ans-1&&aa[i]<=aa[pos]&&bb[i]>=bb[pos]){ans--;pos=i;System.out.print(" "+pos);}}System.out.println();}}}