hdu4521-小明系列问题——小明序列（线段树区间求最值）

 1 #include <set>
 2 #include <map>
 3 #include <cmath>
 4 #include <ctime>
 5 #include <queue>
 6 #include <stack>
 7 #include <cctype>
 8 #include <cstdio>
 9 #include <string>
10 #include <vector>
11 #include <cstdlib>
12 #include <cstring>
13 #include <iostream>
14 #include <algorithm>
15 using namespace std;
16 typedef unsigned long long ull;
17 typedef long long ll;
18 const int inf = 0x3f3f3f3f;
19 const double eps = 1e-8;
20 const int maxn = 1e5+10;
21 int dp[maxn],seg[maxn<<2];        //线段树区间求最值
22 int a[maxn];
23 void update(int l,int r,int pos,int x,int val)
24 {
25     if (l == r)
26     {
27         seg[pos] = val;
28         return ;
29     }
30     int mid = (l + r) >> 1;
31     if (x <= mid)
32         update(l,mid,pos<<1,x,val);
33     if (x > mid)
34         update(mid+1,r,pos<<1|1,x,val);
35     seg[pos] = max(seg[pos<<1],seg[pos<<1|1]);
36 }
37 int query(int l,int r,int pos,int x,int y)
38 {
39     if (x <= l && y >= r)
40         return seg[pos];
41     int mid = (l + r) >> 1;
42     int ans1 = 0,ans2 = 0;
43     if (x <= mid)
44         ans1 = query(l,mid,pos<<1,x,y);
45     if (y > mid)
46         ans2 = query(mid+1,r,pos<<1|1,x,y);
47     return max(ans1,ans2);
48 }
49 int main(void)
50 {
51     #ifndef ONLINE_JUDGE
52         freopen("in.txt","r",stdin);
53     #endif
54     int n,d;
55     while (~scanf ("%d%d",&n,&d))
56     {
57         memset(seg,0,sizeof(seg));
58         int len = 0;
59         for (int i = 0; i < n; i++)
60         {
61             scanf ("%d",a+i);
62             a[i] += 2;                     //因为a[i]可能为0，而我的线段树是从1开始的所以加2
63             len = max(len,a[i]);
64         }
65         int ans = 1;
66         for (int i = 0; i < n; i++)
67         {
68             dp[i] = query(1,len,1,1,a[i]-1) + 1;   //不能等于，所以减1，dp[i]表示以a[i]结尾的最长不降序列长度，
69             if (i >= d)                            //延迟更新，这是这道题的关键，
70                 update(1,len,1,a[i-d],dp[i-d]);
71             ans = max(dp[i],ans);
72         }
73         printf("%d\n",ans);
74     }
75     return 0;
76 }