Subsequence
Time Limit: 1000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 3530
64-bit integer IO format: %I64d Java class name: Main
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
解题：单调队列！马丹，真蛋疼，第一次搞这个。。。。。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cstdlib>
5 #include <vector>
6 #include <climits>
7 #include <algorithm>
8 #include <cmath>
9 #define LL long long
10 #define INF 0x3f3f3f3f
11 using namespace std;
12 const int maxn = 100100;
13 int qa[maxn],qb[maxn],h1,h2,t1,t2;
14 int n,m,k,d[maxn],lst1,lst2;
15 int main(){
16 int i,ans;
17 while(~scanf("%d %d %d",&n,&m,&k)){
18 for(i = 1; i <= n; i++)
19 scanf("%d",d+i);
20 lst2 = lst1 = h1 = h2 = 0;
21 t1 = t2 = -1;
22 ans = 0;
23 for(i = 1; i <= n; i++){
24 while(t1 >= h1 && d[qa[t1]] <= d[i]) t1--;
25 qa[++t1] = i;
26 while(t2 >= h2 && d[qb[t2]] >= d[i]) t2--;
27 qb[++t2] = i;
28 while(d[qa[h1]] - d[qb[h2]] > k){
29 if(qa[h1] < qb[h2]){
30 lst1 = qa[h1++];
31 }else lst2 = qb[h2++];
32 }
33 if(d[qa[h1]] - d[qb[h2]] >= m)
34 ans = max(ans,i-max(lst1,lst2));
35 }
36 printf("%d\n",ans);
37 }
38 return 0;
39 }
40 /*
41 5 0 0
42 1 1 1 1 1
43 5 0 3
44 1 2 3 4 5
45 */
xtu summer individual 5 D - Subsequence,布布扣,bubuko.com