Uva 12295 - RMQ with Shifts

你为何这么叼！

这题说好的只有一组数据呢，还要0和文件结束个蛋？

RMQ with Shifts 

Time limit: 10.000 seconds

 

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

 

---

The Seventh Hunan Collegiate Programming Contest
Problemsetter: Rujia Liu, Special Thanks: Yiming Li & Jane Alam Jan

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define INF 0x7fffffff
#define maxn 111111
int x[maxn<<2];
int mi[maxn<<2];
int tmp[maxn<<2];
int n,m,cnt=0;
void up(int rt){mi[rt]=min(mi[rt<<1],mi[rt<<1|1]);}
void build(int l,int r,int rt){
    if(l==r){scanf("%d",&mi[rt]);tmp[cnt++]=mi[rt];return;}
    int m=(l+r)>>1;
    build(l,m,rt<<1);
    build(m+1,r,rt<<1|1);
    up(rt);
}
void update(int p,int d,int l,int r,int rt){
    if(l==r){mi[rt]=d;return;}
    int m=(l+r)>>1;
    if(p<=m)update(p,d,l,m,rt<<1);
    if(p>m)update(p,d,m+1,r,rt<<1|1);
    up(rt);
}
int query(int L,int R,int l,int r,int rt){
    if(L<=l&&R>=r)return mi[rt];
    int m=(l+r)>>1;
    int res=INF;
    if(L<=m)res=min(res,query(L,R,l,m,rt<<1));
    if(R>m)res=min(res,query(L,R,m+1,r,rt<<1|1));
    return res;
}
int main(){
    while(~scanf("%d%d",&n,&m)&&n){
        build(0,n-1,1);
        char op[31];
        int a,b;
        while(m--){
            scanf("%5s",op);
            if(op[0]=='q'){
                scanf("(%d,%d)",&a,&b);
                a--;b--;
                printf("%d\n",query(a,b,0,n-1,1));
            }
            else{
                int k=0;
                char ch;
                while(ch!='(')ch=getchar();
                while(ch!=')')scanf("%d%c",&x[k++],&ch);
                for(int i=0;i<k;i++)x[i]--;
                int temp=tmp[x[0]];
                for(int i=0;i<k-1;i++)tmp[x[i]]=tmp[x[i+1]];
                tmp[x[k-1]]=temp;
                for(int i=0;i<k;i++)update(x[i],tmp[x[i]],0,n-1,1);
            }
        }
    }
    return 0;
}