線~段~樹

shaoyufei發表於2024-05-08
點選檢視程式碼
#include<bits/stdc++.h>
#define lson id<<1
#define rson id<<1|1
using namespace std;
const int N=1e6+12000;
struct node{
	int l,r,num;
	int ma,mi;
}tr[N<<2];
int a[N];
int n,m;
string str;
int ans=0;
int from,to;
void build(int id,int l,int r){
	tr[id].l=l;
	tr[id].r=r;
	if(l==r){
		tr[id].num=a[l];
		tr[id].ma=a[l];
		tr[id].mi=a[l];
		return;
	}
	int mid=(l+r)/2;
	build(lson,l,mid);
	build(rson,mid+1,r);
	tr[id].num=tr[lson].num+tr[rson].num;
	tr[id].ma=max(tr[lson].ma,tr[rson].ma);
	tr[id].mi=min(tr[lson].mi,tr[rson].mi);
} 
void update(int id,int x,int ad){
	if(tr[id].l==tr[id].r){
		tr[id].num+=ad;
		tr[id].ma+=ad;
		tr[id].mi+=ad;
		return;
	}
	int mid=(tr[id].l+tr[id].r)/2;
	if(x<=mid) update(lson,x,ad);
	else update(rson,x,ad);
	tr[id].num=tr[lson].num+tr[rson].num; 
	tr[id].ma=max(tr[lson].ma,tr[rson].ma);
	tr[id].mi=min(tr[lson].mi,tr[rson].mi);
}
int getsum(int id,int l,int r){
	if(r>=tr[id].r&&l<=tr[id].l){
		return tr[id].num;
	}
	int mid=(tr[id].l+tr[id].r)/2;
	if(mid<l) return getsum(rson,l,r);
	else if(mid>=r) return getsum(lson,l,r);
	else return getsum(lson,l,mid)+getsum(rson,mid+1,r);
}
int getmax(int id,int l,int r){
	if(r>=tr[id].r&&l<=tr[id].l){
		return tr[id].ma;
	}
	int maxn=0;
	int mid=(tr[id].r+tr[id].l)/2;
	if(mid<l) maxn=max(maxn,getmax(rson,l,r));
	else if(mid>=r) maxn=max(maxn,getmax(lson,l,r));
	else maxn=max({maxn,getmax(lson,l,mid),getmax(rson,mid+1,r)});
	return maxn;
}
int getmin(int id,int l,int r){
	if(r>=tr[id].r&&l<=tr[id].l){
		return tr[id].ma;
	}
	int minn=0x7fffffff;
	int mid=(tr[id].l+tr[id].r)/2;
	if(mid<l) minn=min(minn,getmin(rson,l,r));
	else if(mid>=r) minn=min(minn,getmin(lson,l,r));
	else minn=min({minn,getmin(lson,l,mid),getmin(rson,mid+1,r)});
	return minn;
}
int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	cin>>m;
	if(n) build(1,1,n);
	for(int i=1;i<=m;i++){
		cin>>str>>from>>to;
		if(str=="ADD"){
			update(1,from,to);
		}
		else{
			cout<<getsum(1,from,to)<<endl;
		}
	}
}

幾個注意點
1.更新和建樹跑到葉子時要return

if(tr[id].l==tr[id].r){
		tr[id].num+=ad;
		tr[id].ma+=ad;
		tr[id].mi+=ad;
		return;
	}
if(l==r){
		tr[id].num=a[l];
		tr[id].ma=a[l];
		tr[id].mi=a[l];
		return;
	}

2.關於取等

if(mid<l) return getsum(rs,l,r);
	else if(mid>=r) return getsum(ls,l,r);
	else return getsum(ls,l,mid)+getsum(rs,mid+1,r);

image
分別對應以上三行程式碼,因為我們是將l到mid定為lson,mid+1到r定為rson,所以查詢的左邊界要大於mid才能完全在右區間,而右邊界僅需小於等於
3.區間處理(lazy標記)

點選檢視程式碼
#include<bits/stdc++.h>
#define lson id<<1
#define rson id<<1|1
using namespace std;
const int N=1e6+1200;
int a[N];
int n,m;
struct node{
	int l,r,lazy,num;
}tr[N<<2];
void build(int id,int l,int r){
	tr[id].l=l;
	tr[id].r=r;
	if(l==r){
		tr[id].num=a[l];
		return;
	}
	int mid=(l+r)/2;
	build(lson,l,mid);
	build(rson,mid+1,r);
	tr[id].num=tr[lson].num+tr[rson].num;
}
void pushup(int id){
	tr[lson].lazy+=tr[id].lazy;
	tr[rson].lazy+=tr[id].lazy;
	tr[lson].num+=(tr[lson].r-tr[lson].l+1)*tr[id].lazy;
	tr[rson].num+=(tr[rson].r-tr[rson].l+1)*tr[id].lazy;
	tr[id].lazy=0;
}
void update(int id,int l,int r,int ad){
	if(l>tr[id].r||r<tr[id].l){
		return;
	}
	if(tr[id].r<=r&&tr[id].l>=l){
		tr[id].num+=ad*(tr[id].r-tr[id].l+1);
		tr[id].lazy+=ad;
		return;
	}
	int mid=(tr[id].l+tr[id].r)/2;
	pushup(id);
	update(lson,l,r,ad);
	update(rson,l,r,ad);
	tr[id].num=tr[lson].num+tr[rson].num;
}
int getsum(int id,int l,int r){
	if(l>tr[id].r||r<tr[id].l){
		return 0;
	}
	if(tr[id].r<=r&&tr[id].l>=l){
		return tr[id].num;
	}
	int mid=(tr[id].l+tr[id].r)/2;
	pushup(id);
	return getsum(lson,l,mid)+getsum(rson,mid+1,r);
}
int main(){
	string str;
	int from,to,w;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	build(1,1,n);
	cin>>m;
	for(int i=1;i<=m;i++){
		cin>>str;
		if(str=="SUM"){
			cin>>from>>to;
			cout<<getsum(1,from,to)<<endl;
		}
		else {
			cin>>from>>to>>w;
			update(1,from,to,w);
		}
	}
}

我們的目的是,將區間內所有的線段樹都加上對應的值,但問題是,他太多了,所以我們要將新增的要求暫時放一下,所以就有了lazy標記
void pushup(int id){
	tr[lson].lazy+=tr[id].lazy;
	tr[rson].lazy+=tr[id].lazy;//處理id以及它的子樹對應的總和,將子樹以及它的子子樹的處理先放置一邊
	tr[lson].num+=(tr[lson].r-tr[lson].l+1)*tr[id].lazy;
	tr[rson].num+=(tr[rson].r-tr[rson].l+1)*tr[id].lazy;
	tr[id].lazy=0;//處理完清零
}

image
(畫的不是太好,但基本是這麼個意思)

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