UVa 12299 RMQ with Shifts 线段树单点更新

RMQ with Shifts

In the traditional RMQ (Range Minimum Query) problem, we have a static array A. Then for each query (L, R)(L $\le$ 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,..., 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 ( 1 $\le$ n $\le$ 100, 000, 1 $\le$ q $\le$ 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

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

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因为只有交换元素的操作，所以数组里的值是固定的，可以先对num数组做shift处理，然后线段树单点更新求最小值。

scanf("%5s",s);读5个字符到串里。

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#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define N 222222

using namespace std;

const int OO=1e9;

int num[N];

struct Tree
{
    int l;
    int r;
    int min;
} tree[N*4];

void push_up(int root)
{
    tree[root].min=min(tree[root<<1].min,tree[root<<1|1].min);
}

void build(int root,int l,int r)
{
    tree[root].l=l;
    tree[root].r=r;
    if(tree[root].l==tree[root].r)
    {
        tree[root].min=num[l];
        return;
    }
    int mid=(l+r)/2;
    build(root<<1,l,mid);
    build(root<<1|1,mid+1,r);
    push_up(root);
}
void update(int root,int pos,int val)
{
    if(tree[root].l==tree[root].r)
    {
        tree[root].min=val;
        return;
    }
    int mid=(tree[root].l+tree[root].r)/2;
    if(pos<=mid)
        update(root<<1,pos,val);
    else
        update(root<<1|1,pos,val);
    push_up(root);
}
int query(int root,int L,int R)
{
    if(L<=tree[root].l&&R>=tree[root].r)
        return tree[root].min;
    int mid=(tree[root].l+tree[root].r)/2,ret=OO;
    if(L<=mid) ret=min(ret,query(root<<1,L,R));
    if(R>mid) ret=min(ret,query(root<<1|1,L,R));
    return ret;
}

int main()
{
    int n,q,x,y;
    int stk[1111];
    int cnt;
    char s[11];
    while (~scanf("%d%d",&n,&q))
    {
        for (int i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
        }
        build(1,1,n);
        while (q--)
        {
            scanf("%6s",s);
            if (s[0]=='q')
            {
                scanf("%d,%d)",&x,&y);
                printf("%d\n",query(1,x,y));
                //cerr<<x<<" "<<y;
            }
            if (s[0]=='s')
            {
                char c;
                int rm;
                cnt=0;
                while (scanf("%d%c",&x,&c))
                {
                    stk[cnt++]=x;
                    if (cnt==1)
                    {
                        rm=num[stk[0]];
                    }
                    else
                    {
                        num[stk[cnt-2]]=num[stk[cnt-1]];
                        update(1,stk[cnt-2],num[stk[cnt-2]]);
                    }
                    if (c!=',') break;
                }
                num[stk[cnt-1]]=rm;
                update(1,stk[cnt-1],num[stk[cnt-1]]);
            }
        }
    }
    return 0;
}