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 <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--;//递增
while(lx<rx && a[qmax[rx-1]]<a[i])rx--;//递减
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);//因为l从0开始所以i-l不需要+1
}
}
print(ans);
}
return 0;
}