bzoj3188 Upit

题意：

思路：
类似于线段树维护标记。
区间赋值，优先级最高，直接对和进行维护。
区间加等差数列就可以把标记设为 ：首项和公差，这样区间加等差数列的标记就能合并了。
再套平衡树就好了。

ops:
第一道比赛中做的平衡树的题。
因为当时状态比较拉跨，并且区间加等差数列的标记和之前一个线段树加等差数列的平方 的标记搞混了，觉得不能做。知道标记的维护方式，那他就成了裸裸的模板题了。

#include<bits/stdc++.h>
using namespace std;
#define ll long long

const int maxn=300010,INF=1e9;

ll n,m;
struct node{
    
	ll vis[2],v,rnd,siz;
	ll sum;
	ll laz;
	ll fir,sec;
}tree[maxn];

int tot=0,root=0;

void update(int x){
    
	tree[x].siz=tree[tree[x].vis[0]].siz+tree[tree[x].vis[1]].siz+1;
	tree[x].sum=tree[tree[x].vis[0]].sum+tree[tree[x].vis[1]].sum+tree[x].v;
}

void pushdown1(int x){
    
	if(tree[x].laz==0)return;
	if(tree[x].vis[0]){
    
		tree[tree[x].vis[0]].laz=tree[x].laz;
		tree[tree[x].vis[0]].sum=tree[x].laz*tree[tree[x].vis[0]].siz;
		tree[tree[x].vis[0]].v=tree[x].laz;
		tree[tree[x].vis[0]].fir=tree[tree[x].vis[0]].sec=0;
	}
	if(tree[x].vis[1]){
    
		tree[tree[x].vis[1]].laz=tree[x].laz;
		tree[tree[x].vis[1]].sum=tree[x].laz*tree[tree[x].vis[1]].siz;
		tree[tree[x].vis[1]].v=tree[x].laz;
		tree[tree[x].vis[1]].fir=0;
		tree[tree[x].vis[1]].sec=0;
	}
	
	tree[x].laz=0;
}

void pushdown2(int x){
    
	if(tree[x].fir==0)return ;
	if(tree[x].vis[0]){
    
		tree[tree[x].vis[0]].fir+=tree[x].fir;
		tree[tree[x].vis[0]].sec+=tree[x].sec;
		
		int z=tree[tree[x].vis[0]].vis[0];
		tree[tree[x].vis[0]].v+=tree[x].fir+tree[x].sec*(tree[z].siz);
		tree[tree[x].vis[0]].sum+=(2*tree[x].fir+tree[x].sec*(tree[tree[x].vis[0]].siz-1))*tree[tree[x].vis[0]].siz/2;	
	}
	
	if(tree[x].vis[1]){
    
		ll fir=tree[x].fir+tree[x].sec*(tree[tree[x].vis[0]].siz+1);
		tree[tree[x].vis[1]].fir+=fir;
		tree[tree[x].vis[1]].sec+=tree[x].sec;
		
		int z=tree[tree[x].vis[1]].vis[0];
		tree[tree[x].vis[1]].v+=fir+tree[x].sec*(tree[z].siz);
		
		tree[tree[x].vis[1]].sum+=(2*fir+tree[x].sec*(tree[tree[x].vis[1]].siz-1))*tree[tree[x].vis[1]].siz/2;
	}
	
	tree[x].fir=tree[x].sec=0;
}

int newnode(ll v){
    
	tot++;
	tree[tot].siz=1;
	tree[tot].v=v;
	tree[tot].rnd=rand();
	tree[tot].sum=v;
	tree[tot].laz=0;tree[tot].fir=0;
	tree[tot].sec=0;
	return tot;
}
int merge(int x,int y){
    //默认x<y 
	if(!x||!y)return x+y;
	if(tree[x].rnd<tree[y].rnd){
    
		pushdown1(x);
		pushdown2(x);
		tree[x].vis[1]=merge(tree[x].vis[1],y);
		update(x);
		return x;
	}else{
    
		pushdown1(y);
		pushdown2(y);
		tree[y].vis[0]=merge(x,tree[y].vis[0]);
		update(y);
		return y;
	}
}
void split2(int now,int k,ll &x,ll &y){
    //按siz分（找第k大 
	if(!now)x=y=0;
	else {
    
		pushdown1(now);
		pushdown2(now);
		
		if(k<=tree[tree[now].vis[0]].siz)
			y=now,split2(tree[now].vis[0],k,x,tree[now].vis[0]);
		else x=now,split2(tree[now].vis[1],k-tree[tree[now].vis[0]].siz-1,tree[now].vis[1],y);
		update(now);
	}
}
void insert(int v){
    
	ll x,y;
	split2(root,v,x,y);
	ll a;
	cin>>a;
	root=merge(merge(x,newnode(a)),y);
}
int main(){
    
	cin>>n>>m;
	for(int i=1;i<=n;i++)insert(i);
	while(m--){
    
		ll op,x,y,w;
		cin>>op;
		if(op==1){
    
			cin>>x>>y>>w;
			ll a,b,c;
			split2(root,y,a,b);
			split2(a,x-1,a,c);
			tree[c].laz=w;
			tree[c].v=w;
			tree[c].sum=w*tree[c].siz;
			root=merge(merge(a,c),b);
		}else if(op==2){
    
			cin>>x>>y>>w;
			ll a,b,c;
			split2(root,y,a,b);
			split2(a,x-1,a,c);
			tree[c].fir+=w;
			tree[c].v+=w+w*(tree[tree[c].vis[0]].siz);
			tree[c].sec+=w;
			tree[c].sum+=(w+w*(tree[c].siz-1))*tree[c].siz/2;
			root=merge(merge(a,c),b);
		}else if(op==3){
    
			cin>>w;
			insert(w-1);
		}else{
    
			cin>>x>>y;
			ll a,b,c;
			split2(root,y,a,b);
			split2(a,x-1,a,c);
			printf("%lld\n",tree[c].sum);
			root=merge(merge(a,c),b);
		}
	}
	return 0;
}