CSU 1553 Good subsequence（RMQ问题 + 二分）

Posted 2020-09-12 Dh_q

题目链接：http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1553

 

Description

Give you a sequence of n numbers, and a number k you should find the max length of Good subsequence. Good subsequence is a continuous subsequence of the given sequence and its maximum value - minimum value<=k. For example n=5, k=2, the sequence ={5, 4, 2, 3, 1}. The answer is 3, the good subsequence are {4, 2, 3} or {2, 3, 1}.

 

Input

There are several test cases.
Each test case contains two line. the first line are two numbers indicates n and k (1<=n<=10,000, 1<=k<=1,000,000,000). The second line give the sequence of n numbers a[i] (1<=i<=n, 1<=a[i]<=1,000,000,000). 
The input will finish with the end of file.

 

Output

For each the case, output one integer indicates the answer.

 

Sample Input

5 2
5 4 2 3 1
1 1
1

Sample Output

3
1

Hint

Source

 

题意：

　　给你一个n个数，在这些数里面找一段连续区间，这个区间需要满足这个区间里的最大值-最小值 <= k。找出最大的区间长度。

题解：

　　拿到题的时候，就想到RMQ来找区间最值，二分长度找最大值。

　　然后队伍分工合作，一个人写二分，我写RMQ。一不小心写错了一个点wa了一次，QAQ。

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <string>
 6 #include <vector>
 7 #include <map>
 8 #include <set>
 9 #include <queue>
10 #include <sstream>
11 #include <algorithm>
12 using namespace std;
13 #define pb push_back
14 #define mp make_pair
15 #define ms(a, b)  memset((a), (b), sizeof(a))
16 //#define LOCAL
17 #define eps 0.0000001
18 #define LNF (1<<60)
19 typedef long long LL;
20 const int inf = 0x3f3f3f3f;
21 const int maxn = 10000+10;
22 const int mod = 1e9+7;
23 
24 LL st_max[16][maxn], st_min[16][maxn];
25 LL A[maxn];
26 void RMQ_init(int n){
27     for(int i=0;i<n;i++)    st_max[0][i]=st_min[0][i]= A[i];
28     for(int j = 1;(1<<j) <= n;j++){
29         int k = 1<<(j-1);
30         for(int i=0;i+k<n;i++){
31             st_max[j][i] = max(st_max[j-1][i], st_max[j-1][i+k]);
32             st_min[j][i] = min(st_min[j-1][i], st_min[j-1][i+k]);
33         }
34     }
35 }
36 LL RMQ(int a, int b){
37     int dis = abs(b-a) + 1;
38     int k;
39     for(k=0;(1<<k)<=dis;k++);
40     k--;
41 //    printf("%lld %lld\n", max(st_max[k][a], st_max[k][b-(1<<k)+1]), min(st_min[k][a], st_min[k][b-(1<<k)+1]));
42     return max(st_max[k][a], st_max[k][b-(1<<k)+1])-min(st_min[k][a], st_min[k][b-(1<<k)+1]);
43 }
44 int test(int len, int n, int k)
45 {
46     for(int i = len-1; i<n; i++){
47         if(RMQ(i-len+1, i)<=1LL*k)
48             return 1;
49     }
50     return 0;
51 }
52 int main()
53 {
54 #ifdef LOCAL
55     freopen("input.txt", "r", stdin);
56 //      freopen("output.txt", "w", stdout);
57 #endif // LOCAL
58 
59     int n, k;
60     while(~scanf("%d%d", &n, &k)){
61         ms(st_max, 0);
62         ms(st_min, 0);
63         ms(A, 0);
64         for(int i=0;i<n;i++) scanf("%lld", &A[i]);
65         RMQ_init(n);
66 //        printf("%lld\n", RMQ(0, n-1));
67         int l = 0, r = n+1;
68         int mid, ans;
69         while(l<r){
70             mid = (l+r)/2;
71             if(test(mid, n, k)){
72                 ans = mid;
73                 l = mid+1;
74             }
75             else
76                 r = mid;
77         }
78         printf("%d\n", ans);
79     }
80     return 0;
81 }

View Code

 

模板题。XD