xtu summer individual 5 D

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

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
",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 */

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