[dfs序 树状数组] BZOJ 4034 [HAOI2015]T2

两个log可以树链剖分

day2听课时惊闻一个log的做法

但我打low了，并不快...

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#define V G[p].v
using namespace std;
typedef long long ll;

inline char nc()
{
	static char buf[100000],*p1=buf,*p2=buf;
	if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
	return *p1++;
}

inline void read(ll &x)
{
	char c=nc(),b=1;
	for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
	for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

inline void write(ll x)
{
	char s[50],len=0;
	if (x==0) { putchar('0'); return; }
	if (x<0) putchar('-'),x=-x;
	while (x) s[++len]=x%10+'0',x/=10;
	for (int i=len;i;i--) putchar(s[i]);
}

const int N=100005;

#define lowbit(x) ((x)&-(x))

struct BIT{
	ll maxn,c[N];
	inline void init(int n){
		maxn=n;
	}
	inline void add(ll x,ll r){
		for (int i=x;i<=maxn;i+=lowbit(i))
			c[i]+=r;
	}
	inline void add(ll l,ll r,ll x){
		add(l,x); add(r+1,-x);
	}
	inline ll query(ll x){
		ll ret=0;
		for (int i=x;i;i-=lowbit(i))
			ret+=c[i];
		return ret;
	}
}bit1,bit2;

struct edge{
	int u,v,next;
};

edge G[N*2];
int head[N],num;

inline void add(int u,int v,int p)
{
	G[p].u=u; G[p].v=v; G[p].next=head[u]; head[u]=p;
}

ll n,Q,w[N];
int clk,size[N],fat[N],depth[N],tid[N],last[N];

inline int dfs(int u,int fa)
{
	size[u]=1; fat[u]=fa; depth[u]=depth[fa]+1; 
	tid[u]=++clk;
	for (int p=head[u];p;p=G[p].next)
		if (V!=fa)
			size[u]+=dfs(V,u);
	last[u]=tid[u]+size[u]-1;
	return size[u];
}

int main()
{
	ll x,y,order,ans;
	freopen("t.in","r",stdin);
	freopen("t.out","w",stdout);
	read(n); read(Q); bit1.init(n); bit2.init(n);
	for (int i=1;i<=n;i++) read(w[i]);
	for (int i=1;i<n;i++) read(x),read(y),add(x,y,++num),add(y,x,++num);
	dfs(1,0);
	for (int i=1;i<=n;i++) bit1.add(tid[i],last[i],w[i]);
	while (Q--)
	{
		read(order);
		if (order==1)
		{
			read(x); read(y);
			bit1.add(tid[x],last[x],y);
		}
		else if (order==2)
		{
			read(x); read(y);
			bit1.add(tid[x],last[x],-(depth[x]-1)*y);
			bit2.add(tid[x],last[x],y);
		}
		else if (order==3)
		{
			read(x);
			write(bit1.query(tid[x])+bit2.query(tid[x])*depth[x]); putchar('\n');
		}
	}
	return 0;
}