题解 ABC294GDistance Queries on a Tree

Posted 2023-03-21 rui_er 的博客

ABC294G：DFS 序树状数组。

DFS 序树状数组。

不妨以 \\(1\\) 为根，设 \\(\\operatornamedep(u)\\) 表示 \\(u\\) 到根路径的边权和，\\(\\operatornamedis(u,v)\\) 表示 \\(u,v\\) 间路径的边权和，\\(\\operatornameLCA(u,v)\\) 表示 \\(u,v\\) 的最近公共祖先。则显然有：\\(\\operatornamedis(u,v)=\\operatornamedep(u)+\\operatornamedep(v)-2\\times\\operatornamedep(\\operatornameLCA(u,v))\\)。只需要维护每个点的 \\(\\operatornamedep\\)。

给一条边重新赋权值等价于给它的边权加上 \\(\\Delta w\\)，这一操作会导致这条边的子节点所在子树内每个点的 \\(\\operatornamedep\\) 增加 \\(\\Delta w\\)。使用 DFS 序即可将子树转化为区间，区间加、单点查只需要差分后树状数组维护即可。

时间复杂度 \\(\\mathcal O((n+q)\\log n)\\)。

// Problem: G - Distance Queries on a Tree
// Contest: AtCoder - AtCoder Beginner Contest 294
// URL: https://atcoder.jp/contests/abc294/tasks/abc294_g
// Memory Limit: 1024 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x,y,z) for(ll x=(y);x<=(z);x++)
#define per(x,y,z) for(ll x=(y);x>=(z);x--)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) dofreopen(s".in","r",stdin);freopen(s".out","w",stdout);while(false)
#define likely(exp) __builtin_expect(!!(exp), 1)
#define unlikely(exp) __builtin_expect(!!(exp), 0)
using namespace std;
typedef long long ll;

mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
ll randint(ll L, ll R) 
	uniform_int_distribution<ll> dist(L, R);
	return dist(rnd);


template<typename T> void chkmin(T& x, T y) if(x > y) x = y;
template<typename T> void chkmax(T& x, T y) if(x < y) x = y;

const ll N = 2e5+5;

ll n, m, fa[N][19], dis[N], sz[N], bottom[N], dfn[N], tms, val[N];
vector<tuple<ll, ll, ll>> e[N];

struct BIT 
	ll c[N];
	ll lowbit(ll x) return x & (-x);
	void add(ll x, ll k) for(; x <= n; x += lowbit(x)) c[x] += k;
	ll ask(ll x) ll k = 0; for(; x; x -= lowbit(x)) k += c[x]; return k;
bit;

void dfs(ll u, ll f) 
	fa[u][0] = f;
	rep(i, 1, 18) fa[u][i] = fa[fa[u][i-1]][i-1];
	dis[u] = dis[f] + 1;
	dfn[u] = ++tms;
	sz[u] = 1;
	for(auto i : e[u]) 
		ll v, w, id;
		tie(v, w, id) = i;
		if(v != f) 
			dfs(v, u);
			sz[u] += sz[v];
			bottom[id] = v;
			bit.add(dfn[v], w);
			bit.add(dfn[v]+sz[v], -w);
		
	


ll LCA(ll u, ll v) 
	if(dis[u] < dis[v]) swap(u, v);
	per(i, 18, 0) if(dis[fa[u][i]] >= dis[v]) u = fa[u][i];
	if(u == v) return u;
	per(i, 18, 0) if(fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
	return fa[u][0];


int main() 
	scanf("%lld", &n);
	rep(i, 1, n-1) 
		ll u, v, w;
		scanf("%lld%lld%lld", &u, &v, &w);
		e[u].emplace_back(v, w, i);
		e[v].emplace_back(u, w, i);
		val[i] = w;
	
	dfs(1, 0);
	for(scanf("%lld", &m); m; m--) 
		ll op, u, v;
		scanf("%lld%lld%lld", &op, &u, &v);
		if(op == 1) 
			ll bot = bottom[u];
			bit.add(dfn[bot], v-val[u]);
			bit.add(dfn[bot]+sz[bot], val[u]-v);
			val[u] = v;
		
		else 
			ll lca = LCA(u, v);
			printf("%lld\\n", bit.ask(dfn[u])+bit.ask(dfn[v])-2*bit.ask(dfn[lca]));
		
	
	return 0;