hdu 3530 Subsequence 单调队列

Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6130    Accepted Submission(s): 2037

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 01 1 1 1 15 0 31 2 3 4 5

 

Sample Output

54

 

Source

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

 

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题意：给出一个序列，问满足最大元素值-最小元素值在给定区间[m,k]内的子串的最大长度是多少。

分析：1.由于区间范围是定的

           2.从左向右枚举右端点，相应的左端点如果越远，那么Max-Min就越大，如果越近则越小。为了将差值控制在[-∝，k]范围内，先要将上次的左端点右移一段，在第一个满足差值<=k的位置停止。

            此时一定可以保证<=k,如果满足>=m，那么更新答案。否则不必更新，也不必移动左端点，因为左端点左移必定会使之不满足差值<=k的条件，如果右移差值只会减小，同样不能够满足>=m的条件。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])#define mem(a,x)  memset(a,x,sizeof a)#define ysk(x)  (1<<(x))typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;const int maxn=100000    ;int n,L,R;int a[maxn+10],qMax[maxn+10],qMin[maxn+10];int main(){   std::ios::sync_with_stdio(false);   while(cin>>n>>L>>R)   {       for1(i,n)  cin>>a[i];       int ans=0;       int rMax=0,fMax=0,rMin=0,fMin=0;       int le=1;       for(int i=1;i<=n;i++)       {           while(rMax-fMax>=1&& a[qMax[rMax-1] ]<=a[i]   )   rMax--;           qMax[rMax++]=i;           while(rMin-fMin>=1&& a[qMin[rMin-1] ]>=a[i]   )   rMin--;           qMin[rMin++]=i;           while(le<i&&a[qMax[fMax]]-a[qMin[fMin]]>R)           {               if(qMax[fMax]==le) fMax++;               if(qMin[fMin]==le) fMin++;               le++;           }           if( a[qMax[fMax]]-a[qMin[fMin]]>=L )  ans=max(ans,  i-le+1);       }       printf("%d\n",ans);   }   return 0;}