bzoj 3221: Obserbing the tree树上询问 树链剖分+线段树

题目大意:

http://www.lydsy.com/JudgeOnline/problem.php?id=3221

题解

啊呀。。。这是昨天的考试题啊。。。直接就粘了。。
4515: [Sdoi2016]游戏类似,还更简单了一些。
(考场上因为没开long long爆了25分)

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(ll &x){
	x=0;char ch;bool flag = false;
	while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
	while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
const ll maxn = 100010;
struct Edge{
	ll to,next;
}G[maxn<<1];
ll head[maxn],cnt;
void add(ll u,ll v){
	G[++cnt].to = v;
	G[cnt].next = head[u];
	head[u] = cnt;
}
#define v G[i].to
ll fa[maxn],dfn[maxn],top[maxn],dep[maxn],siz[maxn];
ll son[maxn],dfs_clock,seq[maxn],sum[maxn];
void dfs(ll u){
	siz[u] = 1;
	for(ll i = head[u];i;i=G[i].next){
		if(v == fa[u]) continue;
		fa[v] = u;
		dep[v] = dep[u] + 1;
		dfs(v);
		siz[u] += siz[v];
		if(siz[son[u]] < siz[v]) son[u] = v;
	}
}
void dfs(ll u,ll tp){
	top[u] = tp;dfn[u] = ++dfs_clock;
	seq[dfs_clock] = u;
	if(son[u]) dfs(son[u],tp);
	for(ll i = head[u];i;i=G[i].next){
		if(v == son[u] || v == fa[u]) continue;
		dfs(v,v);
	}
}
#undef v
inline ll lca(ll u,ll v){
	while(top[u] != top[v]){
		if(dep[top[u]] < dep[top[v]]) swap(u,v);
		u = fa[top[u]];
	}return dep[u] < dep[v] ? u : v;
}

struct Node{
	Node *ch[2];
	ll sum,k,b;
	ll dep_sum;
	void update(){
		dep_sum = ch[0]->dep_sum + ch[1]->dep_sum;
		sum = ch[0]->sum + ch[1]->sum;
	}
}*null,*root[maxn],*rt;
Node mem[25001001],*it;
inline void init(){
	it = mem;
	null = it++;null->ch[0] = null->ch[1] = null;
	null->sum = null->k = null->b = 0;null->dep_sum = 0;
	root[0] = null;
}
inline Node *newNode(){
	Node *p = it++;p->ch[0] = p->ch[1] = null;
	p->sum = p->k = p->b = p->dep_sum = 0;
	return p;
}
ll L,R,K,B,n,T;
Node *change(Node *rt,ll l,ll r){
	Node *p = newNode();*p = *rt;
	if(L <= l && r <= R){
		p->k += K;p->b += B;
		p->sum += K*p->dep_sum + B*(r-l+1);
		return p;
	}ll mid = (l+r) >> 1;
	if(L <= mid) p->ch[0] = change(p->ch[0],l,mid);
	if(R >  mid) p->ch[1] = change(p->ch[1],mid+1,r);
	p->update();p->sum += p->dep_sum*p->k + (r-l+1)*p->b;
	return p;
}
void change(ll u,ll v,ll a,ll b){
	ll lc = lca(u,v);
	K = -b;B = (a+dep[u]*b);
	while(top[u] != top[lc]){
		L = dfn[top[u]];R = dfn[u];
		rt = change(rt,1,n);
		u = fa[top[u]];
	}
	L = dfn[lc];R = dfn[u];
	rt = change(rt,1,n);
	K = b;B -= 2*dep[lc]*b;
	while(top[v] != top[lc]){
		L = dfn[top[v]];R = dfn[v];
		rt = change(rt,1,n);
		v = fa[top[v]];
	}
	L = dfn[lc];R = dfn[v];
	if(L == R) return;++L;
	rt = change(rt,1,n);
}
ll query(Node *p,ll l,ll r,ll L,ll R){
	if(L <= l && r <= R){
		return p->sum;
	}
	ll ret = (sum[min(r,R)] - sum[max(l,L)-1])*p->k + p->b*(min(r,R) - max(l,L) + 1);
	ll mid = (l+r) >> 1;
	if(L <= mid) ret += query(p->ch[0],l,mid,L,R);
	if(R >  mid) ret += query(p->ch[1],mid+1,r,L,R);
	return ret;
}
inline ll query(ll u,ll v){
	ll ret = 0;
	while(top[u] != top[v]){
		if(dep[top[u]] < dep[top[v]]) swap(u,v);
		ret += query(rt,1,n,dfn[top[u]],dfn[u]);
		u = fa[top[u]];
	}if(dep[u] > dep[v]) swap(u,v);
	ret += query(rt,1,n,dfn[u],dfn[v]);
	return ret;
}
Node *build(ll l,ll r){
	Node *p = newNode();
	if(l == r){
		p->dep_sum = dep[seq[l]];
		p->sum = 0;
		return p;
	}
	ll mid = (l+r) >> 1;
	p->ch[0] = build(l,mid);
	p->ch[1] = build(mid+1,r);
	p->update();return p;
}
char s[10];
int main(){
	init();
	ll last = 0,total = 0;
	ll m;read(n);read(m);
	for(ll i=1,u,v;i<n;++i){
		read(u);read(v);
		add(u,v);add(v,u);
	}dfs(1);dfs(1,1);root[0] = build(1,n);
	sum[0] = 0;
	for(ll i=1;i<=n;++i) sum[i] = sum[i-1] + dep[seq[i]];
	rt = root[0];
	ll u,v,a,b,w;
	while(m--){
		scanf("%s",s);
		if(*s == 'c'){
			read(u);read(v);read(a);read(b);
			u ^= last;v ^= last;
			change(u,v,a,b);
			root[++total] = rt;
		}else if(*s == 'q'){
			read(u);read(v);
			u ^= last;v ^= last;
			printf("%lld
",last = query(u,v));
		}else{
			read(w);w ^= last;
			rt = root[w];
		}
	}
	getchar();getchar();
	return 0;
}
原文地址:https://www.cnblogs.com/Skyminer/p/6446003.html