CSU 1553 Good subsequence（RMQ问题 + 二分）

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。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <string>
 #include <vector>
 #include <map>
 #include <set>
 #include <queue>
 #include <sstream>
 #include <algorithm>
 using namespace std;
 #define pb push_back
 #define mp make_pair
 #define ms(a, b)  memset((a), (b), sizeof(a))
 //#define LOCAL
 #define eps 0.0000001
 #define LNF (1<<60)
 typedef long long LL;
 const int inf = 0x3f3f3f3f;
 const int maxn = +;
 const int mod = 1e9+;

 LL st_max[][maxn], st_min[][maxn];
 LL A[maxn];
 void RMQ_init(int n){
     for(int i=;i<n;i++)    st_max[][i]=st_min[][i]= A[i];
     for(int j = ;(<<j) <= n;j++){
         int k = <<(j-);
         for(int i=;i+k<n;i++){
             st_max[j][i] = max(st_max[j-][i], st_max[j-][i+k]);
             st_min[j][i] = min(st_min[j-][i], st_min[j-][i+k]);
         }
     }
 }
 LL RMQ(int a, int b){
     int dis = abs(b-a) + ;
     int k;
     for(k=;(<<k)<=dis;k++);
     k--;
 //    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]));
     return max(st_max[k][a], st_max[k][b-(<<k)+])-min(st_min[k][a], st_min[k][b-(<<k)+]);
 }
 int test(int len, int n, int k)
 {
     for(int i = len-; i<n; i++){
         if(RMQ(i-len+, i)<=1LL*k)
             return ;
     }
     return ;
 }
 int main()
 {
 #ifdef LOCAL
     freopen("input.txt", "r", stdin);
 //      freopen("output.txt", "w", stdout);
 #endif // LOCAL

     int n, k;
     while(~scanf("%d%d", &n, &k)){
         ms(st_max, );
         ms(st_min, );
         ms(A, );
         for(int i=;i<n;i++) scanf("%lld", &A[i]);
         RMQ_init(n);
 //        printf("%lld\n", RMQ(0, n-1));
         int l = , r = n+;
         int mid, ans;
         while(l<r){
             mid = (l+r)/;
             if(test(mid, n, k)){
                 ans = mid;
                 l = mid+;
             }
             else
                 r = mid;
         }
         printf("%d\n", ans);
     }
     return ;
 }

模板题。XD