HDU-3530 Subsequence(单调队列)

Posted 2020-10-11 可惜没如果=_=

Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7390    Accepted Submission(s): 2498

Problem Description

There is a sequence of integers. Your task is to find the longest subsequence that satisfies the following condition: the difference between the maximum element and the minimum element of the subsequence is no smaller than m and no larger than k.

 

 

Input

There are multiple test cases.
For each test case, the first line has three integers, n, m and k. n is the length of the sequence and is in the range [1, 100000]. m and k are in the range [0, 1000000]. The second line has n integers, which are all in the range [0, 1000000].
Proceed to the end of file.

 

 

Output

For each test case, print the length of the subsequence on a single line.

 

 

Sample Input

5 0 0 1 1 1 1 1 5 0 3 1 2 3 4 5

 

 

Sample Output

5 4

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（10）——Host by HEU

 

 

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有一点突破的迹象qwq  就是求某一个区间的最值并且这个区间的左右端点在变的情况一般都用单调队列 or 单调栈 来做QwQ

dalao勿喷qwq

 1 #include "bits/stdc++.h"
 2 using namespace std;
 3 typedef long long LL;
 4 const int MAX=1e5+5;
 5 int n,m,k;
 6 int a[MAX];
 7 deque <int> q1,q2;
 8 inline int read(){
 9     int an=0,x=1;char c=getchar();
10     while (c<‘0‘ || c>‘9‘) {if (c==‘-‘) x=-1;c=getchar();}
11     while (c>=‘0‘ && c<=‘9‘) {an=an*10+c-‘0‘;c=getchar();}
12     return an*x;
13 }
14 int main(){
15     freopen ("subsequence.in","r",stdin);
16     freopen ("subsequence.out","w",stdout);
17     int i,j;
18     while (~scanf("%d%d%d",&n,&m,&k)){
19         int ans=0,last=0;
20         for (i=1;i<=n;i++) a[i]=read();
21         while (q1.size()) q1.pop_back(); while (q2.size()) q2.pop_back();
22         for (i=1;i<=n;i++){
23             while (q1.size() && a[i]>a[q1.back()]) q1.pop_back();
24             while (q2.size() && a[i]<a[q2.back()]) q2.pop_back();
25             q1.push_back(i),q2.push_back(i);
26             while (q1.size() && q2.size() && (a[q1.front()]-a[q2.front()])>k){
27                 if (q1.front()<q2.front()){
28                     last=q1.front(),q1.pop_front();
29                 }
30                 else if (q1.front()>q2.front()){
31                     last=q2.front(),q2.pop_front();
32                 }
33                 else{
34                     last=q1.front(),q1.pop_front(),q2.pop_front();
35                 }
36             }
37             if (q1.size() && q2.size() && (a[q1.front()]-a[q2.front()])>=m){
38                 ans=max(ans,i-last);
39             }
40         }
41         printf("%d\n",ans);
42     }
43     return 0;
44 }