Codeforces 696E ...Wait for it...（树链剖分）

题目链接  ...Wait for it...

考虑树链剖分。

对于树上的每个点开一个set，记录当前该节点上所有的girls。

每个节点初始的权值为set中的最小值。

询问的时候每次在路径上寻找最小值，并返回这个点的编号。

然后把这个点的set的第一个元素取出，换成下一个元素。

因为女孩总数不超过1e5，所以总查询次数不会超过1e5

修改操作用lazy标记就可以了。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define lson		i << 1, L, mid
#define rson		i << 1 | 1, mid + 1, R
#define ls		i << 1
#define rs		i << 1 | 1
#define MP		make_pair
#define fi		first
#define se		second

typedef long long LL;
typedef pair <LL, int> PII;

const int N  = 1e5 + 10;
const LL inf = 1e18;

int n, m, q;
int id, cnt = 0;
int ret[N];
int at[N], sz[N], son[N], top[N], f[N], in[N], out[N], deep[N], father[N];
vector <int> v[N];
set <PII> s[N];
LL lazy[N << 2];
PII t[N << 2];

void dfs(int x, int fa, int dep){
	sz[x] = 1;
	son[x] = 0;
	father[x] = fa;
	deep[x] = dep;
	for (auto u : v[x]){
		if (u == fa) continue;
		dfs(u, x, dep + 1);
		sz[x] += sz[u];
		if (sz[son[x]] < sz[u]) son[x] = u;
	}
}

void dfs2(int x, int tp){
	top[x] = tp;
	in[x] = f[x]  = ++id;
	if (son[x]) dfs2(son[x], tp);
	for (auto u : v[x]){
		if (u == father[x] || u == son[x]) continue;
		dfs2(u, u);
	}
	out[x] = id;
}

void pushup(int i){
	t[i] = min(t[ls], t[rs]);
}

void pushdown(int i){
	if (lazy[i]){
		lazy[ls] += lazy[i];
		lazy[rs] += lazy[i];
		t[ls].fi += lazy[i];
		t[rs].fi += lazy[i];
		lazy[i] = 0;
	}
}

void build(int i, int L, int R){
	lazy[i] = 0;
	if (L == R){
		t[i] = *s[L].begin();
		return;
	}

	int mid = (L + R) >> 1;
	build(lson);
	build(rson);
	pushup(i);
}

void update(int i, int L , int R , int l, int r, LL val){
	if (l <= L && R <= r){
		lazy[i] += val;
		t[i].fi += val;
		return;
	}

	pushdown(i);
	int mid = (L + R) >> 1 ;
	if (l <= mid) update(lson, l, r, val);
	if (r > mid)  update(rson, l, r, val);
	pushup(i);
}

void modify(int i, int L, int R, int x){
	if (L == R){
		s[L].erase(s[L].begin());
		t[i] = *s[L].begin();
		t[i].fi += lazy[i];
		return;
	}
	pushdown(i);
	int mid = (L + R) >> 1 ;
	if (x <= mid) modify(lson, x) ;
	else modify(rson, x);
	pushup(i);
}

PII query(int i, int L, int R, int l, int r){
	if (l <= L && R <= r) return t[i];
	pushdown(i);
	int mid = (L + R) >> 1 ;
	if (r <= mid) return query(lson, l, r);
	if (l >  mid) return query(rson, l, r);
	return min(query(lson, l, r), query(rson, l, r));
}

int Query(int x, int y){
	PII ans = PII(inf, 0);
	while (top[x] ^ top[y]){
		if (deep[top[x]] < deep[top[y]]) swap(x, y);
		ans = min(ans, query(1, 1, n, f[top[x]], f[x]));
		x = father[top[x]];
	}
	if (deep[x] > deep[y]) swap(x, y);
	ans = min(ans, query(1, 1, n, f[x], f[y]));
	return ans.se;
}

int main(){

	scanf("%d%d%d", &n, &m, &q);
	rep(i, 2, n){
		int x, y;
		scanf("%d%d", &x, &y);
		v[x].push_back(y);
		v[y].push_back(x);
	}

	dfs(1, 0, 0);
	dfs2(1, 1);

	rep(i, 1, n) s[i].insert(MP(inf, 0));
	at[0] = N - 1;

	rep(i, 1, m){
		int x;
		scanf("%d", &x);
		at[i] = x;
		s[f[x]].insert(MP(i, i));
	}

	build(1, 1, n);

	while (q--){
		int op, x, y, lim;
		LL val;
		scanf("%d", &op);
		if (op == 2){
			scanf("%d%lld", &x, &val);
			update(1, 1, n, in[x], out[x], val);
		}
		else{
			scanf("%d%d%d", &x, &y, &lim);
			cnt = 0;
			while (lim--){
				int idx = Query(x, y);
				if (idx == 0) break;
				ret[++cnt] = idx;
				modify(1, 1, n, f[at[idx]]);
			}
			printf("%d", cnt);
			rep(i, 1, cnt) printf(" %d", ret[i]);
			putchar(10);
		}
	}

	return 0;
}