ACM模板

---

区间取最值

Gorgeous Sequence

区间最值操作往往采用以下办法

线段树维护：

• 区间最大值$\text{mx}$
• 区间严格次大值$\text{smx}$
• 区间和$\text{sum}$
• 区间最大值个数$\text{cnt}$
• 区间最值懒标记$\text{lazy}$

实现区间最小值操作，考虑u节点维护的区间，进行如下处理

• 当$\text{mx}\le x$，显然这次修改不会对这个节点维护的区间产生影响，直接退出。
• 当$\text{smx}<x<\text{mx}$，显然这次修改只会影响到这个区间所有的最大值，由此直接根据最大值个数更新区间和并且更新区间最大值并打上懒标记然后退出即可。
• 当$\text{x}\le \text{smx}$，无法直接更新于是递归左右子树。

时间复杂度请直接查阅吉老师2016年国家集训队论文（不太会~

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<set>
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<random>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int mod=998244353;
const int N=1000010;
// 快读
template <class T>
inline void read(T &res)
{res = 0; bool bo = 0; char c;while (((c = getchar()) < '0' || c > '9') && c != '-');if (c == '-') bo = 1; else res = c - 48;while ((c = getchar()) >= '0' && c <= '9')res = (res << 3) + (res << 1) + (c - 48);if (bo) res = ~res + 1;
}
struct node
{int l,r;ll mx,smx,sum;int cnt;ll lazy;
}tree[N<<2];
int n,m;
ll a[N];
void pushup(int u)
{tree[u].sum=tree[u<<1].sum+tree[u<<1|1].sum;if(tree[u<<1].mx>tree[u<<1|1].mx){tree[u].mx=tree[u<<1].mx;tree[u].smx=max(tree[u<<1].smx,tree[u<<1|1].mx);tree[u].cnt=tree[u<<1].cnt;}else if(tree[u<<1].mx<tree[u<<1|1].mx){tree[u].mx=tree[u<<1|1].mx;tree[u].smx=max(tree[u<<1|1].smx,tree[u<<1].mx);tree[u].cnt=tree[u<<1|1].cnt;}else {tree[u].mx=tree[u<<1|1].mx;tree[u].smx=max(tree[u<<1].smx,tree[u<<1|1].smx);tree[u].cnt=tree[u<<1].cnt+tree[u<<1|1].cnt;}
}
void pushlazy(int u,ll x)
{if(tree[u].mx<=x) return; tree[u].sum+=(x-tree[u].mx)*tree[u].cnt;tree[u].mx=tree[u].lazy=x;
}
void pushdown(int u)
{if(tree[u].lazy==-1) return;pushlazy(u<<1,tree[u].lazy),pushlazy(u<<1|1,tree[u].lazy);tree[u].lazy=-1;
}
void build(int u,int l,int r)
{tree[u]={l,r,0,-1,0,0,-1};if(l==r) {tree[u].mx=tree[u].sum=a[l];tree[u].cnt=1;return;}int mid=l+r>>1;build(u<<1,l,mid),build(u<<1|1,mid+1,r);pushup(u);
}
void modify(int u,int l,int r,ll x)
{if(x>=tree[u].mx) return;if(tree[u].l>=l&&tree[u].r<=r&&tree[u].smx<x){pushlazy(u,x);return;}pushdown(u);int mid=tree[u].l+tree[u].r>>1;if(l<=mid) modify(u<<1,l,r,x);if(r>mid) modify(u<<1|1,l,r,x);pushup(u);}
ll qmax(int u,int l,int r)
{if(tree[u].l>=l&&tree[u].r<=r) return tree[u].mx;int mid=tree[u].r+tree[u].l>>1;pushdown(u);ll v=-1;if(l<=mid) v=max(v,qmax(u<<1,l,r));if(r>mid) v=max(v,qmax(u<<1|1,l,r));pushup(u);return v;
}
ll qsum(int u,int l,int r)
{if(tree[u].l>=l&&tree[u].r<=r) return tree[u].sum;int mid=tree[u].r+tree[u].l>>1;pushdown(u);ll v=0;if(l<=mid) v+=qsum(u<<1,l,r);if(r>mid) v+=qsum(u<<1|1,l,r);pushup(u);return v;
}
int main()
{//IO;int T=1;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);for(int i=1;i<=n;i++) scanf("%lld",&a[i]);build(1,1,n);while(m--){int op,l,r;scanf("%d%d%d",&op,&l,&r);if(op==0){ll v;scanf("%lld",&v);modify(1,l,r,v);}else if(op==1)printf("%lld\n",qmax(1,l,r));elseprintf("%lld\n",qsum(1,l,r));}}return 0;
}