CSU1553 Good subsequence —— 二分 + RMQ/线段树

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

Sample Output

3

1

题解：

之前学了RMQ，线段树，树状数组，但是一直不知道他们在哪里能派上用场。通过这题，终于找到他们的用武之地了：区间查询最大最小值。

解决了查询区间最大最小值的问题，剩下的就是二分了。这里是二分长度。

RMQ：

 #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_min[i][] = a[i];

         st_max[i][] = a[i];

     }

     for(int j = ; (<<j)<=n; j++)//枚举长度

     for(int i = ; i+(<<j)-<n; i++)//枚举起点

     {

         st_min[i][j] = min(st_min[i][j-],st_min[i+(<<(j-))][j-]);

         st_max[i][j] = max(st_max[i][j-],st_max[i+(<<(j-))][j-]);

     }

 }

 LL RMQ(int l, int r)//查询

 {

     int k = ;

     while((<<(k+))<=r-l+)

         k++;

     return max(st_max[l][k],st_max[r-(<<k)+][k]) - min(st_min[l][k],st_min[r-(<<k)+][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))

     {

         for(int i=;i<n;i++)

             scanf("%lld", &a[i]);

         ms(st_max, );

         ms(st_min, );

         RMQ_init(n);

         int l = , r = n;

         while(l<=r)

         {

             int mid = (l+r)/;

             if(test(mid, n, k))

                 l = mid+;

             else

                 r = mid-;

         }

         printf("%d\n", r);

     }

     return ;

 }

线段树：

 #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 1000000000000

 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 build(int rt, int l, int r)

 {

     if(l==r)

     {

         st_max[rt] = a[r];

         st_min[rt] = a[r];

         return;

     }

     int mid = (l+r)>>;

     build(rt*,l,mid);

     build(rt*+,mid+,r);

     st_max[rt] = max(st_max[rt*], st_max[rt*+]);

     st_min[rt] = min(st_min[rt*], st_min[rt*+]);

 }

 //由于最大最小值都要查询，而return只能返回一个，所以用ma和mi记录最小值

 LL query(int rt, int l, int r, int x, int y, LL &ma, LL &mi)

 {

     if(x<=l && y>=r)

     {

         ma = max(ma,st_max[rt]);

         mi = min(mi,st_min[rt]);

         return ma - mi;

     }

     int mid = (l+r)>>;

     if(y<=mid) query(rt*,l,mid,x,y,ma,mi);

     else if(x>=mid+) query(rt*+,mid+,r,x,y,ma,mi);

     else query(rt*,l,mid,x,mid,ma,mi),query(rt*+,mid+,r,mid+,y,ma,mi);

     return ma - mi;

 }

 int test(int len, int n, int k)

 {

     for(int i = len-; i<n; i++)

     {

         LL ma = -LNF, mi = LNF;

         if(query(,,n-, i-len+, i, ma,mi)<=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)!=EOF)

     {

         for(int i=;i<n;i++)

             scanf("%lld", &a[i]);

         build(,,n-);

         int l = , r = n;

         while(l<=r)

         {

             int mid = (l+r)/;

             if(test(mid, n,k))

                 l = mid+;

             else

                 r = mid-;

         }

         printf("%d\n", r);

     }

     return ;

 }