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
Recommend zhengfeng   |   We have carefully selected several similar problems for you:  

3535 3529 3528 3527 3415 

題意：

給你一個長度為n的數列，求最長子區間的長度，使得區間的最大值與最小值的差s滿足,
m<=s<=k

思路：

這題很容易想到用兩個單調佇列維護當前最值，
作為判斷條件，如果差值大於k了，就去掉較前面的那個佇列元素，並把區間頭更新為它的標號+1，

這裡注意差值小於m並不需要去掉元素，還有更新答案時要先判斷是否滿足條件才能更新。

程式碼：

#include <bits/stdc++.h>
#define pr(x) cout << #x << "= " << x << "  " ；
#define pl(x) cout << #x << "= " << x << endl;
#define ll __int64
using  namespace  std;

template<class T> 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<class T> 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(lm<rm && a[qmin[rm-1]]>a[i])rm--;//遞增
      while(lx<rx && a[qmax[rx-1]]<a[i])rx--;//遞減
      qmin[rm++]=i;
      qmax[rx++]=i;
      while(a[qmax[lx]]-a[qmin[lm]]>k){
        l=(qmin[lm]<qmax[lx])?qmin[lm++]:qmax[lx++];
      }
      if(a[qmax[lx]]-a[qmin[lm]]>=m){
        ans=max(ans, i-l);//因為l從0開始所以i-l不需要+1
      }
    }
    print(ans);
  }
  return 0;
}