bzoj 2599 [IOI2011]Race 点分治

题面

题目传送门

解法

记\(mn_i\)表示路径长度为\(i\)最少经过多少条边

然后进行点分治

假设当前的根为\(x\)，那么先把\(x\)的子树一个一个访问，访问完一棵子树后更新答案

递归到下面的子树时把影响撤销

时间复杂度：\(O(n\ log\ n)\)

代码

#include <bits/stdc++.h>
#define inf 1 << 30
#define N 1000010
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Edge {
	int next, num, v;
} e[N * 3];
int n, K, ans, cnt, rt, now, vis[N], siz[N], f[N], d[N], s[N], mn[N];
void add(int x, int y, int v) {
	e[++cnt] = (Edge) {e[x].next, y, v};
	e[x].next = cnt;
}
void getr(int x, int fa) {
	siz[x] = 1, f[x] = 0;
	for (int p = e[x].next; p; p = e[p].next) {
		int k = e[p].num, v = e[p].v;
		if (vis[k] || k == fa) continue;
		getr(k, x); siz[x] += siz[k];
		chkmax(f[x], siz[k]);
	}
	chkmax(f[x], now - siz[x]);
	if (f[x] < f[rt]) rt = x;
}
void dfs(int x, int fa) {
	if (d[x] <= K) chkmin(ans, mn[K - d[x]] + s[x]);
	for (int p = e[x].next; p; p = e[p].next) {
		int k = e[p].num, v = e[p].v;
		if (k == fa || vis[k]) continue;
		d[k] = d[x] + v, s[k] = s[x] + 1;
		dfs(k, x);
	}
}
void update(int x, int fa) {
	if (d[x] <= K) chkmin(mn[d[x]], s[x]);
	for (int p = e[x].next; p; p = e[p].next) {
		int k = e[p].num, v = e[p].v;
		if (k == fa || vis[k]) continue;
		update(k, x);
	}
}
void calc(int x, int v) {
	d[x] = v, s[x] = 1;
	dfs(x, 0); update(x, 0);
}
void Clear(int x, int fa) {
	if (d[x] > K) return; mn[d[x]] = inf;
	for (int p = e[x].next; p; p = e[p].next) {
		int k = e[p].num;
		if (vis[k] || k == fa) continue;
		Clear(k, x);
	}
}
void work(int x) {
	vis[x] = 1; mn[0] = 0;
	for (int p = e[x].next; p; p = e[p].next) {
		int k = e[p].num, v = e[p].v;
		if (vis[k]) continue; calc(k, v);
	}
	for (int p = e[x].next; p; p = e[p].next) {
		int k = e[p].num, v = e[p].v;
		if (vis[k]) continue; Clear(k, x);
	}
	for (int p = e[x].next; p; p = e[p].next) {
		int k = e[p].num, v = e[p].v;
		if (vis[k]) continue;
		f[rt = 0] = inf, now = siz[k];
		getr(k, x); work(k);
	}
}
int main() {
	read(n), read(K); cnt = n;
	for (int i = 1; i < n; i++) {
		int x, y, v;
		read(x), read(y), read(v);
		add(++x, ++y, v), add(y, x, v);
	}
	f[rt = 0] = inf, now = n, ans = inf;
	for (int i = 1; i <= K; i++) mn[i] = inf;
	getr(1, 0);
	work(rt);
	if (ans == inf) cout << "-1\n";
		else cout << ans << "\n";
	return 0;
}