RMQ with Shifts （线段树）

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
shif t(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 shif t(2, 4, 5, 7) yields {6, 8, 4, 5, 4, 1, 2}. After that,
shif t(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 ≤ 250, 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

题目大概：

给你n个数，m条询问。每条询问，可以是查询区间最小值，或者是改变某些值的位置，即把第二个数的值放在第一个，第三个数放在第二个，第一个放在最后一个。形成循环，相当于单点更新。

思路：

直接用线段树维护区间最小值就行了，不过输入比较麻烦。

代码：

#include 

using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn=1e5+10;
const int INF=0x3f3f3f3f;
int sum[maxn<<2];
int a[maxn];
int b[maxn];
int cnt=0;
char q[100];
int get_b(int st,int en){
    int cnt=0;
    int tm,tmp=0;
    for(int i=st;i='0'&&q[i]<='9'){
           tm=q[i]-'0';
           tmp*=10;
           tmp+=tm;
        }else{
            b[cnt++]=tmp;
            tmp=0;
        }
    }
    return cnt;
}
void pushup(int rt)
{
    sum[rt]=min(sum[rt<<1],sum[rt<<1|1]);
}
void build(int l,int r,int rt)
{
    if(l==r)
    {
        scanf("%d",&sum[rt]);
        a[++cnt]=sum[rt];
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
void update(int p,int sc,int l,int r,int rt)
{

    if(l==r)
    {
        sum[rt]=sc;
        return;
    }
    int m=(l+r)>>1;
    if(p<=m)update(p,sc,lson);
    else update(p,sc,rson);
    pushup(rt);
}

int quert(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        return sum[rt];

    }
    int m=(l+r)>>1;
    int ret=INF;
    if(L<=m)ret=min(ret,quert(L,R,lson));
    if(R>m)ret=min(ret,quert(L,R,rson));

    return ret;
}


int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    build(1,n,1);
    while(m--)
    {
        scanf("%s",q);
        int l=strlen(q);
        int le=get_b(6,l);
        if(q[0]=='s')
        {
            int pre=b[0];
            int ans1=a[pre];
            for(int i=1;i