TZOJ 4325 RMQ with Shifts(线段树查询最小，暴力更新)

描述

In the traditional RMQ (Range Minimum Query) problem, we have a static array A. Then for each query (L, R) (L<=R), we report the minimum value among A[L], A[L+1], …, A[R]. Note that the indices start from 1, i.e. the left-most element is A[1].

In this problem, the array A is no longer static: we need to support another operation shift(i1, i2, i3, …, ik) (i1<i2<...<ik, k>1): we do a left “circular shift” of A[i1], A[i2], …, A[ik].

For example, if A={6, 2, 4, 8, 5, 1, 4}, then shift(2, 4, 5, 7) yields {6, 8, 4, 5, 4, 1, 2}. After that, shift(1,2) yields {8, 6, 4, 5, 4, 1, 2}.

输入

There will be only one test case, beginning with two integers n, q (1<=n<=100,000, 1<=q<=120,000), the number of integers in array A, and the number of operations. The next line contains n positive integers not greater than 100,000, the initial elements in array A. Each of the next q lines contains an operation. Each operation is formatted as a string having no more than 30 characters, with no space characters inside. All operations are guaranteed to be valid. Warning: The dataset is large, better to use faster I/O methods.

输出

For each query, print the minimum value (rather than index) in the requested range.

样例输入

7 5
6 2 4 8 5 1 4
query(3,7)
shift(2,4,5,7)
query(1,4)
shift(1,2)
query(2,2)

样例输出

1
4
6

题意

给你N个数，有两个操作

1.查询区间最小

2.给定几个位置，循环左移

题解

可以看出循环左移的区间不是很大，那么直接拿线段树暴力更新

代码

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define ll long long
 5 
 6 const int maxn=1e5+5;
 7 
 8 int n,q;
 9 int Min[maxn<<2];
10 
11 void build(int l,int r,int rt)
12 {
13     if(l==r)
14     {
15         scanf("%d",&Min[rt]);
16         return;
17     }
18     int mid=(l+r)>>1;
19     build(l,mid,rt<<1);
20     build(mid+1,r,rt<<1|1);
21     Min[rt]=min(Min[rt<<1],Min[rt<<1|1]);
22 }
23 
24 void update(int L,int c,int l,int r,int rt)
25 {
26     if(l==r)
27     {
28         Min[rt]=c;
29         return;
30     }
31     int mid=(l+r)>>1;
32     if(L<=mid)update(L,c,l,mid,rt<<1);
33     else update(L,c,mid+1,r,rt<<1|1);
34     Min[rt]=min(Min[rt<<1],Min[rt<<1|1]);
35 }
36 
37 int query(int L,int R,int l,int r,int rt)
38 {
39     if(L<=l&&r<=R)
40         return Min[rt];
41     int mid=(l+r)>>1,ans=0x3f3f3f3f;
42     if(L<=mid)ans=min(ans,query(L,R,l,mid,rt<<1));
43     if(R>mid)ans=min(ans,query(L,R,mid+1,r,rt<<1|1));
44     return ans;
45 }
46 
47 int main()
48 {
49     scanf("%d%d",&n,&q);
50     build(1,n,1);
51     for(int i=0;i<q;i++)
52     {
53         getchar();
54         int pre,cur;
55         if(getchar()=='s')
56         {
57             while(getchar()!='(');
58             scanf("%d",&pre);
59             int first=query(pre,pre,1,n,1);
60             while(getchar()!=')')
61             {
62                 scanf("%d",&cur),
63                 update(pre,query(cur,cur,1,n,1),1,n,1);
64                 pre=cur;
65             }
66             update(pre,first,1,n,1);
67         }
68         else
69         {
70             while(getchar()!='(');
71             scanf("%d,%d)",&pre,&cur);
72             printf("%d
",query(pre,cur,1,n,1));
73         }
74     }
75     return 0;
76 }