RMQ with Shifts（未解决的题目）

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RMQ with Shifts

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 5   Accepted Submission(s) : 2

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Problem Description

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}.

Input

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.

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 

数组越界的代码：
#include <iostream>
#include <cstdio>
#include <string>
#include <cstdio>

using namespace std;
#define Min(x,y)  x<y?x:y

struct node
{
    int l,r;
    int min;
}sg_tree[1000000];

int f[1000100];

void make(int l1,int r1,int num)
{
    if(l1<r1)
    {
        int mid=(l1+r1)>>1;
        sg_tree[num].l=l1;
        sg_tree[num].r=r1;
        make(l1,mid,2*num);
        make(mid+1,r1,2*num+1);
        sg_tree[num].min=Min(sg_tree[2*num].min,sg_tree[2*num+1].min);
    }
    else if(l1==r1)
    {
        sg_tree[num].l=l1,sg_tree[num].r=r1;
        sg_tree[num].min=f[l1];
    }
}

void insert(int num,int index,int d)
{
    int mid=(sg_tree[num].l+sg_tree[num].r)>>1;
    if(sg_tree[num].l==sg_tree[num].r)
      {
          sg_tree[num].min=d;
          return ;
      }
    if(index<=mid)
    {
        insert(2*num,index,d);
        sg_tree[num].min=Min(sg_tree[2*num].min,sg_tree[2*num+1].min);
    }
    else
    {
        insert(2*num+1,index,d);
        sg_tree[num].min=Min(sg_tree[2*num].min,sg_tree[2*num+1].min);
    }
}
int query(int l,int r,int num)
{
    int mid=(sg_tree[num].l+sg_tree[num].r)>>1;
    if(l<=sg_tree[num].l&&r>=sg_tree[num].r)
      return sg_tree[num].min;
    else if(r<=mid)
      return query(l,r,2*num);
    else if(l>mid)
      return query(l,r,2*num+1);
    else
      return Min(query(l,mid,2*num),query(mid+1,r,2*num+1));
}

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++)
         scanf("%d",&f[i]);
         make(1,n,1);
         while(m--)
         {
            char cmd[330];
            scanf("%s",cmd);
          // puts(cmd);
            int l,r;
            if(cmd[0]=='q')
            {
                l=cmd[6]-'0';
                r=cmd[8]-'0';
                //cout<<l<< "  " <<r<<endl;
                cout<<query(l,r,1)<<endl;;
            }
           else
           {
              int temp=0,array[300];
              for(int i=6;cmd[i];i+=2)
              {
                  array[temp++]=cmd[i]-'0';
              }
             // cout<<array[0]<< " " ;
             int pre=f[array[0]];
              for(int i=0;i<temp-1;i++)
              {
                  insert(1,array[i],f[array[i+1]]);
                  f[array[i]]=f[array[i+1]];
              }
              insert(1,array[temp-1],pre);
              f[array[temp-1]]=pre;
           }
         }
    }
    return 0;
}