xtu summer individual 5 D - Subsequence

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。。。。。

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <vector>
 #include <climits>
 #include <algorithm>
 #include <cmath>
 #define LL long long
 #define INF 0x3f3f3f3f
 using namespace std;
 const int maxn = ;
 int qa[maxn],qb[maxn],h1,h2,t1,t2;
 int n,m,k,d[maxn],lst1,lst2;
 int main(){
     int i,ans;
     while(~scanf("%d %d %d",&n,&m,&k)){
         for(i = ; i <= n; i++)
             scanf("%d",d+i);
         lst2 = lst1 = h1 = h2 = ;
         t1 = t2 = -;
         ans = ;
         for(i = ; i <= n; i++){
             while(t1 >= h1 && d[qa[t1]] <= d[i]) t1--;
             qa[++t1] = i;
             while(t2 >= h2 && d[qb[t2]] >= d[i]) t2--;
             qb[++t2] = i;
             while(d[qa[h1]] - d[qb[h2]] > k){
                 if(qa[h1] < qb[h2]){
                     lst1 = qa[h1++];
                 }else lst2 = qb[h2++];
             }
             if(d[qa[h1]] - d[qb[h2]] >= m)
                 ans = max(ans,i-max(lst1,lst2));
         }
         printf("%d\n",ans);
     }
     return ;
 }
 /*
 5 0 0
 1 1 1 1 1
 5 0 3
 1 2 3 4 5
 */