RMQ with Shifts(线段树)

RMQ with Shifts

Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u

Practice NBUT 1113

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

Hint

无

//线段树的应用，不是很难，写出来还是对线段树有更深一点的了解的

#include <stdio.h>
#include <string.h>

#define MAXN 100005

struct Node
{
    int min;
    int l,r;
}node[3*MAXN];//节点
int pos[MAXN];//记录叶节点的位置
int num[MAXN];//记录最开始的数
int shift[50];//记录shift里的数

int Min(int a,int b)
{
    return a<b?a:b;
}

int Build(int left,int right,int k)
{
    node[k].l=left;
    node[k].r=right;
    if (left==right)//到叶节点
    {
        node[k].min=num[left];
        pos[left]=k;
        return node[k].min;
    }
    int mid=(left+right)/2;

    node[k].min=Min(Build(left ,mid,2*k),Build(mid+1,right,2*k+1));
    return node[k].min;
}

int Query(int left,int right,int k)
{
    if (left==node[k].l&&right==node[k].r)
    {
        return node[k].min;
    }
    int mid=(node[k].l+node[k].r)/2;
    if (left>mid) return Query(left,right,2*k+1);
    else if (right<=mid) return Query(left,right,2*k);
    return Min(Query(left,mid,2*k),Query(mid+1,right,2*k+1));
}

void Update(int k)
{
    k/=2;
    while (k!=0)
    {
        node[k].min=Min(node[2*k].min,node[2*k+1].min);
        k/=2;
    }
}

int Get_shift(char str[])
{
    int i,j;
    int n=0;
    int len=strlen(str);
    for (i=6;i<len;i++)
    {
        int temp=0;
        for (j=i;str[j]!=','&&str[j]!=')';j++)
        {
            temp+=str[j]-'0';
            temp*=10;
        }
        temp/=10;
        shift[++n]=temp;
        i=j;
    }
    return n;
}

int main()
{
    int n,m;
    int i,j;
    scanf("%d%d",&n,&m);
    for (i=1;i<=n;i++)
        scanf("%d",&num[i]);

    Build(1,n,1);              //递归建树
    char str[50];
    int left,right;

    for (i=1;i<=m;i++)
    {
        scanf("%s",&str);
        if (str[0]=='q')
        {
            left=0,right=0;
            for (j=6;str[j]!=',';j++)
            {
                left+=str[j]-'0';
                left*=10;
            }
            left/=10;
            for (j++;str[j]!=')';j++)
            {
                right+=str[j]-'0';
                right*=10;
            }
            right/=10;
            printf("%d
",Query(left,right,1));//查找区间内最小的
        }
        if (str[0]=='s')
        {
            int shift_num=Get_shift(str);//获得shift里面的数

            int temp=node[pos[shift[1]]].min;
            for (j=2;j<=shift_num;j++)
                node[pos[shift[j-1]]].min=node[pos[shift[j]]].min;
            node[pos[shift[j-1]]].min=temp;
            
            for (j=1;j<=shift_num;j++)
                Update(pos[shift[j]]);
        }

    }
    return 0;
}