（TLE待AC）CSU 1110 RMQ with Shifts

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本文探讨了一个涉及RMQ（区间最小值查询）问题的编程挑战，其中包括数组元素的动态更新及区间查询操作。文章通过一个具体的编程实例介绍了如何使用线段树解决这一问题，并分享了作者在实现过程中遇到的超时问题及其初步解决方案。

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1110: RMQ with Shifts

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 298 Solved: 66
[ Submit][ Status][ Web Board]

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

Source

湖南省第七届大学生计算机程序设计竞赛

对线段树的理解还是不够深入。

这题的需要的是RMQ区间更新。每次操作还要左移序列，和以前写的更新函数不太一样。我在左移操作之后又Build了一下，虽然运行结果是对的，但是最终还是超时了。

现在先把超时的代码存档。待我再研究研究。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define INF 0x3f3f3f3f
#define N 100005
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
int a[N];
int mi[N<<2];
void PushUp(int rt)
{
        mi[rt]=min(mi[rt<<1],mi[rt<<1|1]);
}
void Build(int l,int r,int rt)
{
        if(l==r)
        {
                mi[rt]=a[l];
                return;
        }
        int m=(l+r)>>1;
        Build(lson);
        Build(rson);
        PushUp(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 ret=INF;
        if(L<=m)
                ret=min(ret,query(L,R,lson));
        if(R>m)
                ret=min(ret,query(L,R,rson));
        return ret;
}
int main()
{
        int n,q;
        while(scanf("%d%d",&n,&q)>0)
        {
                for(int i=1;i<=n;i++)
                        scanf("%d",&a[i]);
                Build(1,n,1);
                while(q--)
                {
                        char s[30];
                        scanf("%s",s);
                        if(s[0]=='q')
                        {
                                int a,b;
                                sscanf(s+5,"(%d,%d)",&a,&b);
                                printf("%d\n",query(a,b,1,n,1));
                        }
                        else
                        {
                                int shift[30],k=0,len=strlen(s);
                                for(int i=6;i<len;i+=2)
                                        shift[k++]=(s[i]-'0');
                                int temp=a[shift[0]];
                                for(int i=0;i<k-1;i++)
                                        a[shift[i]]=a[shift[i+1]];
                                a[shift[k-1]]=temp;
                                Build(1,n,1);
                        }
                }
        }
        return 0;
}

/**************************************************************
    Problem: 1110
    User: 0905130123
    Language: C++
    Result: Time Limit Exceed
****************************************************************/