Subsequence（hdu3530）

Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6141    Accepted Submission(s): 2041

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

思路：优先队列+尺取；

我们不用去管这个子串中的最大最小的距离是否大于等于m，我们只要保证这个值小于等于k时继续向右端扩展，应为向右端扩展尺取时，当前值是越来越大的。比如当[l,r]满足dis>=m，那么[l,r+s],的dis>=m;所以我们不需要管m。然后就是尺取中，更新最大最小值的问题，这个用优先队列维护下。复杂度N*log(N);

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<stdlib.h>
  6 #include<queue>
  7 #include<stack>
  8 using namespace std;
  9 typedef long long LL;
 10 int ask[100005];
 11 int cnt[100005];
 12 struct node1
 13 {
 14     int x;
 15     int id;
 16     bool operator<(const node1 &cx)const
 17     {
 18         if(cx.x == x)
 19             return cx.id<id;
 20         else return cx.x>x;
 21     }
 22 };
 23 struct node2
 24 {
 25     int x;
 26     int id;
 27     bool operator<(const node2 &cx)const
 28     {
 29         if(cx.x == x)
 30             return cx.id<id;
 31         else return cx.x<x;
 32     }
 33 };
 34 priority_queue<node1>que1;
 35 priority_queue<node2>que2;
 36 int main(void)
 37 {
 38     int n,m,k;
 39     while(scanf("%d %d %d",&n,&m,&k)!=EOF)
 40     {
 41         while(!que1.empty())que1.pop();
 42         while(!que2.empty())que2.pop();
 43         int i,j;
 44         for(i = 0; i < n; i++)
 45         {
 46             scanf("%d",&ask[i]);
 47         }
 48         int l = 0;
 49         int r = 0;
 50         int cc = 0;
 51         int ma = ask[0];
 52         int mi = ask[0];
 53         int x = abs(ma-mi);
 54         if(x <= k&&x >= m)cc = 1;
 55         node1 ak;
 56         node2 ap;
 57         ak.x =ask[0];
 58         ak.id = 0;
 59         ap.x = ask[0];
 60         ap.id = 0;
 61         que1.push(ak);
 62         que2.push(ap);
 63         while(l<=r&&r<n)
 64         {
 65             while(x <= k &&r < n-1)
 66             {
 67                 r++;
 68                 int c = abs(ask[r]-ma);
 69                 c = max(abs(ask[r]-mi),c);
 70                 if(c > k)
 71                 {   //printf("%d %d\n",l,r);
 72                     r--;
 73                     break;
 74                 }
 75                 node1 ac;
 76                 ac.x = ask[r];
 77                 ac.id = r;
 78                 node2 bc;
 79                 bc.x= ask[r];
 80                 bc.id = r;
 81                 que1.push(ac);
 82                 que2.push(bc);
 83                 if(ask[r] > ma)
 84                 {
 85                     ma = ask[r];
 86                 }
 87                 else if(ask[r] < mi)
 88                 {
 89                     mi = ask[r];
 90                 }
 91                 x = abs(ma-mi);//printf("%d\n",x);
 92             }
 93             if(x >= m)
 94             {
 95                 cc = max(cc,r-l+1);
 96             }
 97             if(ask[l] == ma)
 98             {
 99                 while(!que1.empty())
100                 {
101                     node1 acc = que1.top();
102                     if(acc.id <= l)
103                     {
104                         que1.pop();
105                     }
106                     else
107                     {
108                         ma = acc.x;
109                         break;
110                     }
111                 }
112             }
113             if(ask[l]==mi)
114             {
115                 while(!que2.empty())
116                 {
117                     node2 acc = que2.top();
118                     if(acc.id <= l)
119                     {
120                         que2.pop();
121                     }
122                     else
123                     {
124                         mi = acc.x;
125                         break;
126                     }
127                 }
128             }
129             l++;
130             if(l == r+1)
131             {   //printf("%d\n",r);
132                 r++;
133                 node1 akk;
134                 node2 app;
135                 akk.x =ask[r];
136                 akk.id = r;
137                 app.x = ask[r];
138                 app.id = r;
139                 que1.push(akk);
140                 que2.push(app);
141                 mi = ask[r];
142                 ma = ask[r];
143             }
144         }
145         printf("%d\n",cc);
146     }
147     return 0;
148 }