NYOJ 1012 RMQ with Shifts （线段树）

题目链接

• In the traditional RMQ (Range Minimum Query) problem, we have a static array A. Then for each query (L, R)(L \leR), 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.
  
  输入
  
  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 100 characters, with no space characters inside. All operations are guaranteed to be valid.
  
  输出
  
  For each query, print the minimum value (rather than index) in the requested range.
  
  样例输入
  
  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)
  
  样例输出
  
  1
  
  4
  
  6

分析：

对于一个下标从1到n的序列，主要有两种操作，一种是query(查询)某个区间内的最小值，另一种是shift(循环左移一位)就是将这些下标对应的值循环向左移动一位，当然第一位的值会移动到最后一位。我们首先肯定是将前一位的值更改为后一位，但是在更改最后一位的时候第一位的值已经改变了，所以应该将第一位的值提前存储下来。

代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<stack>
#include<math.h>
using namespace std;
int a[100009];
struct Node
{
    int left;///左区间
    int right;///右区间
    int num;///该区间内的最小值
} node[400009];

void build(int root,int le,int ri)
{

    node[root].left=le;
    node[root].right=ri;
    if(le==ri)///左区间与右区间相等，相当于找到了某个特定的值
    {
        node[root].num=a[le];
        return ;
    }
    int mid=(le+ri)/2;
    build(root*2,le,mid);
    build(root*2+1,mid+1,ri);
    node[root].num=min(node[root*2].num,node[root*2+1].num);
}

void Update(int root,int n1,int n2)
{

    if(node[root].left==n1&&node[root].right==n1)///先找到这个数字，然后把这个值更新了
    {
        node[root].num=n2;
        return;
    }
    if(n1<=node[root*2].right)///该节点在左子树中
        Update(root*2,n1,n2);
    else if(n1>=node[root*2+1].left)///节点在右子树中
        Update(root*2+1,n1,n2);
    node[root].num=min(node[root*2].num,node[root*2+1].num);///取左右子树的较小值
}

int Query(int root,int le,int ri)
{
    if(node[root].left==le&&node[root].right==ri)///找到特定的区间，返回值
    {
        return node[root].num;
    }
    else if(node[root*2].right>=ri)///区间在左子树中
    {
        return  Query(root*2,le,ri);
    }
    else if(node[root*2+1].left<=le)///区间在右子树中
    {
        return Query(root*2+1,le,ri);
    }
    else
    {
        int mid=(node[root].left+node[root].right)/2;///左右子树中都有
        int num1=Query(root*2,le,mid);
        int num2=Query(root*2+1,mid+1,ri);
        return min(num1,num2);
    }
}

int main()
{
    //freopen("1.txt","r",stdin);
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1; i<=n; i++)
        scanf("%d",&a[i]);
    char ch[109];
    build(1,1,n);
    while(m--)
    {
        int b[1000];
        int j=0;
        int sum=0;
        scanf(" %s",ch);
        for(int i=6; ch[i]!='\0'; i++)///将每个数字提取出来
        {
            if(ch[i]>='0'&&ch[i]<='9')
            {
                sum=sum*10+ch[i]-'0';
            }
            else
            {
                if(ch[i]==','||ch[i]==')')
                {
                    b[j++]=sum;
                    sum=0;
                }
            }
        }
        if(ch[0]=='q')
        {
            printf("%d\n",Query(1,b[0],b[1]));
        }
        else
        {
            int NUM=a[b[0]];///把第一位保存下来
            for(int i=0; i<j-1; i++)///先把a数组更新了，再把树进行更新
                a[b[i]]=a[b[i+1]];
            a[b[j-1]]=NUM;
            for(int i=0; i<j-1; i++)
            {
                Update(1,b[i],a[b[i]]);
            }
            Update(1,b[j-1],NUM);
        }
    }
    return 0;
}