Range Minimum Queries

最新推荐文章于 2024-03-20 16:35:01 发布

UMR小豪

最新推荐文章于 2024-03-20 16:35:01 发布

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分类专栏：线段树文章标签：线段树思维

本文链接：https://blog.youkuaiyun.com/qq_33362864/article/details/73647930

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这篇博客探讨了如何处理Range Minimum Queries的问题，利用线段树进行优化。文章通过一个例子展示了在处理0型查询（查找指定范围内最小值）和1型查询（对指定范围内的元素进行位与赋值操作）时，如何有效地更新线段树。关键在于将问题转化为二进制表示，只更新那些在当前状态下为0的位，从而减少不必要的计算。

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You are given an array of N integers. You should support the following queries on this array.

0 L R : Find the minimum integer in the range A_L, A_L+1, ..., A_R.
1 L R X : You should apply the assignment A[i] = A[i] & X, for all indices i in range [L, R], where & denotes bitwise AND operation.

Input

First line of the input contains two space separated integers N and Q.

Second line contains N integer numbers denoting array A.

In the next Q lines, each contain one of the queries described above.

Output

For each query of the type 0, output a single line containing the answer of the query.

Constraints

1 ≤ N, Q ≤ 10⁵
1 ≤ A_i, X ≤ 10⁹
1 ≤ L ≤ R ≤ N

Example

中文题意请看这里：

https://odzkskevi.qnssl.com/e9034bc038dcbe22b7b4894298e69c38?v=1497576044

线段树。重点在于更新上。一开始只是简单的想当前值为0，就没必要更新了。想的太过简单了。再深入一步想，应该转化成二进制，看哪一位上是0，就没有必要再更新了。这样就能得到更大的优化，因为每个点最多更新30次。

用一个变量state来记录当前区间的二进制状态。更新的时候，如果要更新的当前状态位上是0，则返回。

剩下的看代码吧。

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>

using namespace std;
const int MAXN = 1e5+7;
int n,m;
struct node
{
    int l,r;
    int MIN;
    int state;
}tree[MAXN<<2];
void push_up(int i)
{
    tree[i].state = tree[i<<1].state | tree[i<<1|1].state;
    tree[i].MIN = tree[i<<1].MIN<tree[i<<1|1].MIN?tree[i<<1].MIN:tree[i<<1|1].MIN;
}
void build_tree(int i,int l,int r)
{
    tree[i].l = l ;
    tree[i].r = r;
    if(l == r)
    {
        scanf("%d",&tree[i].MIN);
        tree[i].state = tree[i].MIN;
        return ;
    }
    int mid = (l+r)>>1;
    build_tree(i<<1,l,mid);
    build_tree(i<<1|1,mid+1,r);
    push_up(i);
}
void updata(int i,int l,int r,int x)
{
    if((tree[i].state & x) == 0)return;
    if(tree[i].l == tree[i].r)
    {
        tree[i].MIN -= x;
        tree[i].state -= x;
        return ;
    }
    int mid = (tree[i].l+tree[i].r)>>1;
    if(r <= mid)updata(i<<1,l,r,x);
    else if(l > mid)updata(i<<1|1,l,r,x);
    else
    {
        updata(i<<1,l,mid,x);
        updata(i<<1|1,mid+1,r,x);
    }
    push_up(i);
}
int ask(int i,int l,int r)
{
    if(tree[i].l == l && tree[i].r == r)return tree[i].MIN;
    int mid = (tree[i].l + tree[i].r)>>1;
    if(r <= mid)return ask(i<<1,l,r);
    else if(l > mid)return ask(i<<1|1,l,r);
    else return min(ask(i<<1,l,mid),ask(i<<1|1,mid+1,r));

}


int main()
{
    scanf("%d%d",&n,&m);
    build_tree(1,1,n);
    int x,l,r,y;
    while(m--)
    {
        scanf("%d%d%d",&x,&l,&r);
        if(!x)printf("%d\n",ask(1,l,r));
        else
        {
            scanf("%d",&y);
            int t = 1;
            for(int i = 0 ; i < 30 ; ++i)
            {
                if((y & 1) == 0)updata(1,l,r,t);
                y >>= 1;
                t <<= 1;
            }
        }
    }
    return 0;
}