Educational Codeforces Round 20 G. Periodic RMQ Problem（线段树动态开点）

题目链接：
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题意：
给你一个n个数的序列，将它复制粘贴k次得到新的序列。
在新的序列上有2<script type="math/tex" id="MathJax-Element-142">2</script>个操作：区间赋值，区间取最小值。

题解：其实就是线段树动态开点。其他和普通的线段树差不多吧。

代码：

#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
typedef long long ll;
#define lson id << 1, l, mid  
#define rson id << 1 | 1, mid, r 

//int minn[123456*4];
//int lazy[123456*4];
struct segment_tree
{
    int n;
    //  cout<<"minn.size="<<minn.size()<<endl;
//  cout<<"lazy.size="<<lazy.size()<<endl;
    vector<int>a;
    vector<int>minn;
    vector<int>lazy;
    segment_tree(const vector<int> &v):a(v)
    {
        n=a.size();
        minn.resize(n*4);
        lazy.resize(n*4);
        build(1,0,n);
    }
//  cout<<"minn.size="<<minn.size()<<endl;
//  cout<<"lazy.size="<<lazy.size()<<endl;
    void pushup(int id)
    {
        minn[id]=min(minn[id<<1],minn[id<<1|1]);
    }

    void build(int id,int l,int r)
    {
        lazy[id] = 0;
        if(r-l<=1)
        {
            //printf("minn[%d]=%d\n",id,minn[id]);
            minn[id]=a[l];
            return;
        }
        int mid=(l+r)>>1;
        build(lson);
        build(rson);
        pushup(id);
    }
    void shift(int id) 
    {
        if (lazy[id])
        {
            minn[id<<1]=lazy[id<<1]=minn[id<<1|1]=lazy[id<<1|1]=lazy[id];
            lazy[id] = 0;
        }
    }
    void update(int x, int y, int v, int id, int l, int r)
    {
        if (x>=r || l>=y) return;
        if (x<=l && r<=y)
        {
            minn[id] = lazy[id] = v;
            return;
        }
        shift(id);
        int mid = (l+r)>>1;
        update(x, y, v, lson);
        update(x, y, v, rson);
        pushup(id);
    }
    int query(int x,int y,int id,int l,int r)
    {
        if(x>=r || l>=y)return 2e9;
        if(x<=l && r<=y) 
        {
            return minn[id];
        }
        shift(id);
        int mid=(l+r)>>1;
        int ans=min(query(x,y,lson),query(x,y,rson));
        return ans;
    }
};

int l[123456], r[123456];
int op[123456];
int val[123456];
//vector<int>aa;
vector<int>a;
vector<int>b;
/*
3 1
1 2 3
3
2 1 3
1 1 2 4
2 1 3

1
3
*/
int main()
{
    //freopen("in.txt","r",stdin);
    int n,k;
    cin>>n>>k;
    int limit=2e9;
    for(int i=0;i<n;i++)
    {
        int x;
        cin>>x;
        a.push_back(x);
        limit=min(limit,x);
    }
    segment_tree seg(a);
    vector<int>V;
    V.push_back(0);
    V.push_back(n*k);
    int q;
    cin>>q;
    for(int i=0;i<q;i++)
    {
        cin>>op[i]>>l[i]>>r[i];
        l[i]--;
        if(op[i]==1)
        {
            cin>>val[i];
        }
        V.push_back(l[i]);
        V.push_back(r[i]);
    }
    sort(V.begin(),V.end());
    V.erase(unique(V.begin(),V.end()),V.end());

    for(int i=0;i<V.size()-1;i++)
    {
        int left = V[i];
        int right = V[i+1];
        if(right - left >= n)
        {
            b.push_back(limit);
        }
        else
        {
            right %= n;
            if(right==0)
            {
                right = n;
            }
            left %= n;
            if(left < right)
            {
                int res=seg.query(left,right,1,0,n);
                b.push_back(res);
            }
            else
            {
                int res=min(seg.query(left,n,1,0,n), seg.query(0,right,1,0,n));
                b.push_back( res );
            }
        }
    }   
    segment_tree segmt(b);
    for (int i = 0; i < q; i++)
    {
        l[i] = lower_bound(V.begin(), V.end(), l[i]) - V.begin();
        r[i] = lower_bound(V.begin(), V.end(), r[i]) - V.begin();

        if (op[i]==1)
        {
            segmt.update(l[i], r[i], val[i], 1, 0, b.size());   
        }       
        else if(op[i]==2)
        {
            int ans=segmt.query(l[i], r[i], 1, 0, b.size());
            printf("%d\n",ans);
        }

  }
    return 0;
}