C. RMQ with Shifts

C. RMQ with Shifts

Time Limit: 1000ms

Case Time Limit: 1000ms

Memory Limit: 131072KB

 

64-bit integer IO format: %lld      Java class name: Main

 

 

In the traditional RMQ (Range Minimum Query) problem, we have a static array A. Then for each query (L, R) (LR), 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,..., i_k)(i₁ < i₂ < ... < i_k, k > 1)

we do a left ``circular shift" of A[i₁], A[i₂], ..., A[i_k].

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.

Input

There will be only one test case, beginning with two integers n, q ( 1n100, 000, 1q250, 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.

Output

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

Sample Input

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)

Sample Output

1
4
6

解题：RMQ问题，更新比较有新意。。。。。。。。。。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <vector>
 #include <climits>
 #include <algorithm>
 #include <cmath>
 #define LL long long
 using namespace std;
 const int maxn = ;
 struct node{
     int lt,rt,minVal;
 }tree[maxn<<];
 int d[maxn],u[],cnt;
 void build(int lt,int rt,int v){
     tree[v].lt = lt;
     tree[v].rt = rt;
     if(lt == rt){
         tree[v].minVal = d[lt];
         return;
     }
     int mid = (lt+rt)>>;
     build(lt,mid,v<<);
     build(mid+,rt,v<<|);
     tree[v].minVal = min(tree[v<<].minVal,tree[v<<|].minVal);
 }
 int query(int lt,int rt,int v){
     if(tree[v].lt == lt && tree[v].rt == rt) return tree[v].minVal;
     int mid = (tree[v].lt+tree[v].rt)>>;
     if(rt <= mid) return query(lt,rt,v<<);
     else if(lt > mid) return query(lt,rt,v<<|);
     else return min(query(lt,mid,v<<),query(mid+,rt,v<<|));
 }
 void update(int lt,int rt,int v){
     if(tree[v].lt == tree[v].rt){
         tree[v].minVal = d[tree[v].lt];
         return;
     }
     int mid = (tree[v].lt+tree[v].rt)>>;
     if(u[rt] <= mid) update(lt,rt,v<<);
     else if(u[lt] > mid) update(lt,rt,v<<|);
     else{
         int i;
         for(i = lt; u[i] <= mid; i++);
         update(lt,i-,v<<);
         update(i,rt,v<<|);
     }
     tree[v].minVal = min(tree[v<<].minVal,tree[v<<|].minVal);
 }
 int main(){
     int n,m,i,j,len,temp;
     char str[];
     while(~scanf("%d%d",&n,&m)){
         for(i = ; i <= n; i++)
             scanf("%d",d+i);
             build(,n,);
         for(i = ; i < m; i++){
             scanf("%s",str);
             len = strlen(str);
             for(cnt = j = ; j < len;){
                 if(str[j] < '' || str[j] > '') {j++;continue;}
                 temp = ;
                 while(j < len && str[j] >= '' && str[j] <= '') {temp = temp* + (str[j]-'');j++;}
                 u[cnt++] = temp;
             }
             if(str[] == 'q'){
                 printf("%d\n",query(u[],u[],));
             }else{
                 temp = d[u[]];
                 for(cnt--,j = ; j < cnt; j++)
                     d[u[j]] = d[u[j+]];
                 d[u[j]]  = temp;
                 update(,cnt,);
             }
         }
     }
     return ;
 }