UVA 12299 RMQ with Shifts（线段树+点更新）

RMQ with Shifts

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

解题思路：

题目大意：

给你一个数组A是可变的，支持shift(i1,i2,...,ik),表示把元素A[i1],A[i2],....,A[ik],循环左移一位。query(L,R);是询问区间[L,R]的最小值。

算法思想：

比较简单的线段树点更新问题，把循环的更改的值依次单点更新即可。

AC代码：

#include <iostream>
#include <cstdio>
#include <cctype>
#include <cstring>
using namespace std;

const int N = 100005;
struct node{
    int l,r,minn;
}tree[N<<2];
int val[N],tmp[50],sum;

void pushUp(int id){
    tree[id].minn = min(tree[id<<1].minn,tree[id<<1|1].minn);
}

void build(int id,int l,int r){
    tree[id].l = l;
    tree[id].r = r;
    if(tree[id].l == tree[id].r){
        tree[id].minn = val[l];
        return ;
    }
    int mid = (tree[id].l + tree[id].r)>>1;
    build(id<<1,l,mid);
    build(id<<1|1,mid+1,r);
    pushUp(id);
}

void update(int id,int x,int val){
    if(tree[id].l == tree[id].r){
        tree[id].minn = val;
        return ;
    }
    int mid = (tree[id].l+tree[id].r)>>1;
    if(x <= mid)
        update(id<<1,x,val);
    else
        update(id<<1|1,x,val);
    pushUp(id);
}

int query(int m,int l,int r){
    if(tree[m].l  == l && tree[m].r == r)
        return tree[m].minn;
    int mid = (tree[m].l+tree[m].r)>>1;
    if(r <= mid)
        return query(m<<1,l,r);
    if(l > mid)
        return query((m<<1)+1,l,r);
    return min(query(m<<1,l,mid), query((m<<1)+1,mid+1,r));
}

void solve(){
    char str[50];
    int i,j,t;
    scanf("%s",str);
    int len = strlen(str);
    if(str[0] == 's'){
        for(j = 0,i = 0; i < len; ++i){
            if(!isdigit(str[i]))
                continue;
            int t = 0;
            while(isdigit(str[i])){
                t = t*10+(str[i]-'0');
                ++i;
            }
            tmp[j++] = t;
        }
        sum = j;
        t = val[tmp[0]];
        for(i = 0; i < sum-1; ++i){
            val[tmp[i]] = val[tmp[i+1]];
            update(1,tmp[i],val[tmp[i]]);
        }
        val[tmp[i]] = t;
        update(1,tmp[i],val[tmp[i]]);
    }else{
        for(j = 0,i = 0; i < len; ++i){
            if(!isdigit(str[i]))
                continue;
            t = 0;
            while(isdigit(str[i])){
                t = t*10+(str[i]-'0');
                ++i;
            }
            tmp[j++] = t;
        }
        printf("%d\n",query(1,tmp[0],tmp[1]));
    }
}

int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        for(int i = 1; i <= n; ++i)
            scanf("%d",&val[i]);
        build(1,1,n);
        while(m--)
            solve();
    }
    return 0;
}