【HDU3530】【单调队列（双）】Subsequence 【长度为n的数列，求最长子区间的长度，使得区间的最大值与最小值的差满足一个范围】

传送门：HDU 3530 Subsequence

描述：

Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6143    Accepted Submission(s): 2042

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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题意：

给你一个长度为n的数列，求最长子区间的长度，使得区间的最大值与最小值的差s满足,
m<=s<=k

思路：

这题很容易想到用两个单调队列维护当前最值，
作为判断条件，如果差值大于k了，就去掉较前面的那个队列元素，并把区间头更新为它的标号+1，

这里注意差值小于m并不需要去掉元素，还有更新答案时要先判断是否满足条件才能更新。

代码：

#include 
#define pr(x) cout << #x << "= " << x << "  " ；
#define pl(x) cout << #x << "= " << x << endl;
#define ll __int64
using  namespace  std;

template void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const int N=1e5+10;

int n,m,k,a[N];
int qmin[N],qmax[N];

int  main(){
  #ifndef ONLINE_JUDGE
  freopen("in.txt","r",stdin);
  #endif

  while(~scanf("%d%d%d",&n,&m,&k)){
    int lm=0,rm=0,lx=0,rx=0,ans=0,l=0;
    for(int i=1; i<=n; i++){
      read(a[i]);
      while(lma[i])rm--;//递增
      while(lxk){
        l=(qmin[lm]=m){
        ans=max(ans, i-l);//因为l从0开始所以i-l不需要+1
      }
    }
    print(ans);
  }
  return 0;
}