Description

　　給定一個序列A[i]，每次詢問l,r，求[l,r]內最長子串，使得該子串為不上升子串或不下降子串

Input

　　第一行n，表示A數組有多少元素
　　接下來一行為n個整數A[i]
　　接下來一個整數Q，表示詢問數量
　　接下來Q行，每行2個整數l，r

Output

　　對於每個詢問，求[l,r]內最長子串，使得該子串為不上升子串或不下降子串

Sample Input

Sample Output

HINT

　　N,Q<=50000

Solution

　　直接上線段樹，記錄一下一個區間內：左邊最長，右邊最長，整體最優。再分一下不下降和不上升即可。

Code

 1 #include<iostream>    
 2 #include<string>    
 3 #include<algorithm>    
 4
 
 #include<cstdio>    
 5 #include<cstring>    
 6 #include<cstdlib>
 7 #include<cmath>
 8 #include<queue>
 9 using namespace std;  
10 
11 const int ONE = 500005;
12 const int INF = 2147483640;
13 
14 int n,Q;
15 int x,y;
16 int a[ONE];
17 
18 struct power
19 {
 
20         int len;
21         int Lval, Rval;
22         int Lup,Ldn, Rup,Rdn;
23         int Maxup, Maxdn;
24         
25         friend power operator +(power a, power b)
26         {
27             power A;
28             A.len = a.len + b.len;
29             A.Lval = a.Lval;    A.Rval = b.Rval;
30             
31             A.Lup = a.Lup;    if(a.Lup == a.len && a.Rval <= b.Lval) A.Lup += b.Lup;
32             A.Ldn = a.Ldn;    if(a.Ldn == a.len && a.Rval >= b.Lval) A.Ldn += b.Ldn;
33             
34             A.Rup = b.Rup;    if(b.Rup == b.len && a.Rval <= b.Lval) A.Rup += a.Rup;
35             A.Rdn = b.Rdn;    if(b.Rdn == b.len && a.Rval >= b.Lval) A.Rdn += a.Rdn;
36             
37             A.Maxup = max(a.Maxup, b.Maxup);
38             A.Maxdn = max(a.Maxdn, b.Maxdn);
39             
40             if(a.Rval <= b.Lval) A.Maxup = max(A.Maxup, a.Rup+b.Lup);
41             if(a.Rval >= b.Lval) A.Maxdn = max(A.Maxdn, a.Rdn+b.Ldn);
42             
43             return A;
44         }
45 }Node[ONE];
46 
47 int get()
48 {    
49         int res=1,Q=1;char c;    
50         while( (c=getchar())<48 || c>57 ) 
51         if(c==‘-‘)Q=-1; 
52         res=c-48;     
53         while( (c=getchar())>=48 && c<=57 )    
54         res=res*10+c-48;    
55         return res*Q;
56 }
57 
58 void Build(int i,int l,int r)
59 {
60         if(l == r)
61         {
62             Node[i].Lval = Node[i].Rval = a[l];
63             Node[i].len = 1;
64             Node[i].Lup = Node[i].Rup = Node[i].Ldn = Node[i].Rdn = 1;
65             Node[i].Maxup = Node[i].Maxdn = 1;
66             return; 
67         }
68         
69         int mid = l+r>>1;
70         Build(i<<1, l,mid);    Build(i<<1|1, mid+1,r);
71         Node[i] = Node[i<<1] + Node[i<<1|1];
72 }
73 
74 power Query(int i,int l,int r,int L,int R)
75 {
76         if(L==l && r==R) return Node[i];
77         int mid = l+r>>1;
78         if(mid+1 > R) return Query(i<<1, l,mid,L,R);
79         else if(L > mid) return Query(i<<1|1, mid+1,r,L,R);
80         else return Query(i<<1, l,mid,L,mid) + Query(i<<1|1, mid+1,r,mid+1,R); 
81 }
82 
83 int main()  
84 {
85         n = get();
86         for(int i=1; i<=n; i++) a[i] = get();
87         Build(1,1,n);
88         
89         Q = get();
90         while(Q--)
91         {
92             x = get();    y = get();
93             power res = Query(1,1,n,x,y);
94             printf("%d\n", max(res.Maxup, res.Maxdn) );
95         }
96 }

【BZOJ4491】我也不知道題目名字是什麽 [線段樹]