bzoj3011 [Usaco2012 Dec]Running Away From the Barn 左偏树

题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=3011

题解

复习一下左偏树板子。

看完题目就知道是左偏树了。

结果这个板子还调了好久。

大概已经不会写左偏树了。


时间复杂度 \(O(n\log n)\)

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {
	int f = 0, c;
	while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
	x = c & 15;
	while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
	f ? x = -x : 0;
}

const int N = 200000 + 7;

#define lc c[0]
#define rc c[1]

int n; ll l;
int ans[N], rt[N];
ll dis[N];

struct Edge { int to, ne; ll w; } g[N << 1]; int head[N], tot;
inline void addedge(int x, int y, ll z) { g[++tot].to = y, g[tot].w = z, g[tot].ne = head[x], head[x] = tot; }
inline void adde(int x, int y, ll z) { addedge(x, y, z), addedge(y, x, z); }

struct Node { int c[2], dis, s; ll v; } t[N];

inline int merge(int o, int p) {
	if (!o || !p) return o ^ p;
	if (t[o].v < t[p].v) std::swap(o, p);
	t[o].rc = merge(t[o].rc, p);
	if (t[t[o].lc].dis < t[t[o].rc].dis) std::swap(t[o].lc, t[o].rc);
	t[o].dis = t[t[o].rc].dis + 1, t[o].s = t[t[o].lc].s + t[t[o].rc].s + 1;
	return o;
}
inline void pop(int &o) { o = merge(t[o].lc, t[o].rc); }

inline void dfs(int x, int fa = 0) {
	rt[x] = x, t[x].v = dis[x], t[x].s = 1;
	for fec(i, x, y) if (y != fa) dis[y] = dis[x] + g[i].w, dfs(y, x), rt[x] = merge(rt[x], rt[y]);
	while (t[rt[x]].v > dis[x] + l) pop(rt[x]);
	ans[x] = t[rt[x]].s;
}

inline void work() {
	for (int i = 1; i <= n; ++i) if (!dis[i]) dfs(i);
	for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
}

inline void init() {
	read(n), read(l);
	int x; ll y;
	for (int i = 2; i <= n; ++i) read(x), read(y), adde(x, i, y);
}

int main() {
#ifdef hzhkk
	freopen("hkk.in", "r", stdin);
#endif
	init();
	work();
	fclose(stdin), fclose(stdout);
	return 0;
}
posted @ 2019-10-24 21:31  hankeke  阅读(110)  评论(0编辑  收藏  举报