[ZJOI 2018]历史

题意:给定一棵树和点的\(Access\)次数，求切换链的最大值。

考虑修改时实边与虚边的贡献，用\(LCT\)维护此树。

// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2000010;
const int INF = 0x7fffffff;
#define int long long
inline int read()
{
	int q=0,f=1;char ch=getchar();
	while(!isdigit(ch)){
		if(ch=='-') f=-1;ch=getchar();
	}
	while(isdigit(ch)){
		q=q*10+ch-'0';ch=getchar();
	}
	return q*f;
}
int head[maxn];
int cnt;
struct edge
{
	int nxt;
	int to;
}e[maxn<<1];
inline void add(int u,int v){
	e[++cnt].to = v;
	e[cnt].nxt = head[u];
	head[u] = cnt;
	return;
}

int ans;

struct LCT{
	int ch[maxn][2];
	int fa[maxn];
	int stack[maxn];
	int val[maxn];
	int sum[maxn];
	int isum[maxn];
	inline bool isroot(int x){
		return ch[fa[x]][0] != x && ch[fa[x]][1] != x;
	}
	inline void push_up(int now){
		sum[now] = val[now] + sum[ch[now][0]]+sum[ch[now][1]]+isum[now];
	}
	inline void rotate(int x){
		int f=fa[x];
		int ff = fa[f];
		int y = (ch[f][0] == x);
		bool flag = isroot(f);
		fa[x] = ff;
		fa[f] = x;
		fa[ch[x][y]] = f;
		if(!flag) ch[ff][ch[ff][1]==f] = x;
		ch[f][!y] = ch[x][y];
		ch[x][y] = f;
		push_up(f);
	}
	inline void splay(int x){
		while(!isroot(x)){
			int f = fa[x];
			if(!isroot(f)){
				if((ch[fa[f]][0] == f)^(ch[f][0] == x)) rotate(x);
				else rotate(f);
			}
			rotate(x);
		}
		push_up(x);
	}
	inline void access(int x,int v,int nt){
		for(;x;nt = x,x = fa[x]){
			splay(x);
			int res = sum[ch[x][1]] + val[x] + isum[x];
			if(ch[x][1]){
				ans -=((res-sum[ch[x][1]]) << 1);
			}
			else if(res + 1 <= val[x] * 2){
				ans -= 2*(res - val[x]);
			}
			else ans -= (res - 1);
			sum[x] += v;
			isum[x] += v;
			res += v;
			if(res+1 > sum[ch[x][1]]*2){
				isum[x] += sum[ch[x][1]];
				ch[x][1] = 0;
			}
			if(res + 1<= sum[nt]*2){
				ch[x][1] = nt;
				isum[x] -= sum[ch[x][1]];
			}
			if(ch[x][1]){
				ans += 2*(res - sum[ch[x][1]]);
			}
			else if(res + 1 <= val[x]*2){
				ans += 2*(res - val[x]);
			}
			else ans += (res - 1);
		}
	}
	inline void update(int x,int f){
		splay(x);
		int res = sum[ch[x][1]] + val[x] + isum[x];
		if(ch[x][1]) ans -= 2*(res - sum[ch[x][1]]);
		else if(res + 1 <= val[x] * 2){
			ans -= 2*(res - val[x]);
		}
		else ans -= (res - 1);
		val[x] += f;
		sum[x] += f;
		res += f;
		if(res + 1 > sum[ch[x][1]] * 2){
			isum[x] += sum[ch[x][1]];
			ch[x][1] = 0;
		}
		if(ch[x][1]){
			ans += 2*(res - sum[ch[x][1]]);
		}
		else if(res + 1 <= val[x] * 2){
			ans += (res - val[x]) * 2;
		}
		else ans += (res - 1);
		access(fa[x],f,x);
	}
	inline void dfs(int x,int f){
		fa[x] = f;
		sum[x] = val[x];
		int maxm = val[x];
		int i;
		int tmp = x;
		for(int i = head[x];i;i=e[i].nxt){
			int y = e[i].to;
			if(y == f) continue;
			dfs(y,x);
			sum[x] += sum[y];
			if(sum[y] > maxm){
				tmp = y;
				maxm = sum[y];
			}
		}
		ans += min(sum[x]-1,2*(sum[x]-maxm));
		if(tmp != x && sum[x] + 1 <= maxm * 2) ch[x][1] = tmp;
		isum[x] = sum[x] - val[x] - sum[ch[x][1]];
	}
	inline void ins(){
		dfs(1,0);
	}
}lct;

signed main()
{
	int n = read(),m=read();
	for(int i = 1;i <= n; ++i){
		lct.val[i] = read();
	}
	for(int i = 1;i < n; ++i){
		int u=read(),v=read();
		add(u,v);
		add(v,u);
	}
	lct.ins();
	cout<<ans<<endl;
	for(int i = 1;i <=m; ++i){
		int x = read(),v = read();
		lct.update(x,v);
		cout<<ans<<endl;	
	}
	return 0;
	
}