BZOJ 3083 遥远的国度 （换根讨论 + 树链剖分）

题目链接  BZOJ3083

换根不能真正地换。

令当前的根为$cnt$，要查找的子树根为$x$

$1$、$x = cnt$，那么要查找的区域就是整棵树。

$2$、$x$在以$cnt$为根的子树内，那么要查找的区域就是以$x$为根的子树。

$3$、$x$在以$cnt$为根的子树外

　　（1）$x$不是$cnt$的祖先，那么要查找的区域就是以$x$为根的子树。

　　（2）$x$是$cnt$的祖先，设$y$为$x$到$cnt$方向上的第一个点，那么要查找的区域就是整棵树减去以$y$为根的子树。

 

#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	ls		i << 1
#define	rs		i << 1 | 1
#define	mid		((L + R) >> 1)
#define lson		i << 1, L, mid
#define rson		i << 1 | 1, mid + 1, R

const int N = 1e5 + 10;

int f[N], fp[N], son[N], deep[N], father[N], sz[N], a[N], top[N];
int n, m, tot = 0;
int cnt;
int s[N << 2], lazy[N << 2];
vector <int> v[N];

void dfs1(int x, int fa, int dep){
	deep[x] = dep;
	father[x] = fa;
	son[x] = 0;
	sz[x] = 1;
	int ct = (int)v[x].size();
	rep(i, 0, ct - 1){
		int u = v[x][i];
		if (u == fa) continue;
		dfs1(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;
	f[x] = ++tot;
	fp[f[x]] = x;
	if (son[x]) dfs2(son[x], tp);
	int ct = (int)v[x].size();
	rep(i, 0, ct - 1){
		int u = v[x][i];
		if (u == father[x] || u == son[x]) continue;
		dfs2(u, u);
	}
}

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

inline void pushdown(int i){
	if (lazy[i]){
		s[ls] = lazy[ls] = lazy[i];
		s[rs] = lazy[rs] = lazy[i];
		lazy[i] = 0;
	}
}

void update(int i, int L, int R, int l, int r, int val){
	if (l <= L && R <= r){
		s[i] = lazy[i] = val;
		return;
	}

	pushdown(i);
	if (l <= mid) update(lson, l, r, val);
	if (r >  mid) update(rson, l, r, val);
	pushup(i);
}

int query(int i, int L, int R, int l, int r){
	if (l <= L && R <= r) return s[i];

	pushdown(i);
	int ret = INT_MAX;
	if (l <= mid) ret = min(ret, query(lson, l, r));
	if (r >  mid) ret = min(ret, query(rson, l, r));
	return ret;
}

void modify(int x, int y, int val){
	int f1 = top[x], f2 = top[y];
	for (; f1 ^ f2; ){
		if (deep[f1] < deep[f2]) swap(f1, f2), swap(x, y);
		update(1, 1, n, f[f1], f[x], val);
		x = father[f1], f1 = top[x];
	}

	if (x == y){
		update(1, 1, n, f[x], f[y], val);
		return;
	}

	if (deep[x] > deep[y]) swap(x, y);
	update(1, 1, n, f[x], f[y], val);
	return;

}

int twz(int x, int y){
	int t;
	for (; top[x] ^ top[y]; ) t = top[y], y = father[top[y]];
	return x == y ? t : son[x];
}

int LCA(int x, int y){
	for (; top[x] ^ top[y]; ){
		if (deep[top[x]] < deep[top[y]]) swap(x, y);
		x = father[top[x]];
	}

	return deep[x] > deep[y] ? y : x;
}

int main(){

	scanf("%d%d", &n, &m);

	rep(i, 0, 4e5 + 10) s[i] = INT_MAX;

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

	dfs1(1, 0, 1);
	dfs2(1, 1);

	rep(i, 1, n){
		int x;
		scanf("%d", &x);
		update(1, 1, n, f[i], f[i], x);
	}

	scanf("%d", &cnt);

	while (m--){
		int op;
		scanf("%d", &op);
		if (op == 1) scanf("%d", &cnt);
		else if (op == 2){
			int x, y, z;
			scanf("%d%d%d", &x, &y, &z);
			modify(x, y, z);
		}

		else{
			int x;
			scanf("%d", &x);
			if (x == cnt){
				printf("%d\n", query(1, 1, n, 1, n));
				continue;
			}

			int lca = LCA(x, cnt);
			if (lca == x){
				int y = twz(x, cnt);
				int ret = INT_MAX;
				int xx  = f[y], yy = f[y] + sz[y] - 1;
				if (xx >= 1) ret = min(ret, query(1, 1, n, 1, xx - 1));
				if (yy <  n) ret = min(ret, query(1, 1, n, yy + 1, n));
				printf("%d\n", ret);
				continue;
			}

			printf("%d\n", query(1, 1, n, f[x], f[x] + sz[x] - 1));
		}
	}

	return 0;
}