[BZOJ4317]Atm的树

试题描述

Atm有一段时间在虐qtree的题目，于是，他满脑子都是tree，tree，tree……

于是，一天晚上他梦到自己被关在了一个有根树中，每条路径都有边权，一个神秘的声音告诉他，每个点到其他的点有一个距离（什么是距离不用说吧），他需要对于每个点回答：从这个点出发的第k小距离是多少；

如果atm不能回答出来，那么明天4019的闹钟将不会响，4019全寝可能就迟到了，所以atm希望你帮帮他。

输入

第一行，两个正整数n，k，表示树的点数，询问的是第几小距离；

第二~n行，每行三个正整数x，y，w，表示x和y之间有一条边，x为父亲，边权为w；

输出

n行， 每行一个数，第i行输出从i开始第k小距离

输入示例

5 2
1 5 2
1 2 4
2 3 6
2 4 5

输出示例

4
5
10
9
6

数据规模及约定

100% n<=15000, 边权在1~10之间，为了方便，保证1为根；K<=5000

题解

参见这道题。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
	if(Head == Tail) {
		int l = fread(buffer, 1, BufferSize, stdin);
		Tail = (Head = buffer) + l;
	}
	return *Head++;
}
int read() {
	int x = 0, f = 1; char c = Getchar();
	while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = Getchar(); }
	while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = Getchar(); }
	return x * f;
}

#define maxn 15010
#define maxm 30010
#define maxlog 15

int n, m, head[maxn], nxt[maxm], to[maxm], dist[maxm];

void AddEdge(int a, int b, int c) {
	to[++m] = b; dist[m] = c; nxt[m] = head[a]; head[a] = m;
	swap(a, b);
	to[++m] = b; dist[m] = c; nxt[m] = head[a]; head[a] = m;
	return ;
}

int mnd[maxlog][maxn<<1], Log[maxn<<1], clo, dep[maxn], dfn[maxn];
void build(int u, int pa) {
	mnd[0][dfn[u] = ++clo] = dep[u];
	for(int e = head[u]; e; e = nxt[e]) if(to[e] != pa)
		dep[to[e]] = dep[u] + dist[e], build(to[e], u), mnd[0][++clo] = dep[u];
	return ;
}
void rmq_init() {
	Log[1] = 0;
	for(int i = 2; i <= clo; i++) Log[i] = Log[i>>1] + 1;
	for(int j = 1; (1 << j) <= clo; j++)
		for(int i = 1; i + (1 << j) - 1 <= clo; i++)
			mnd[j][i] = min(mnd[j-1][i], mnd[j-1][i+(1<<j-1)]);
	return ;
}
int cdist(int a, int b) {
	int l = dfn[a], r = dfn[b];
	if(l > r) swap(l, r);
	int t = Log[r-l+1];
	return dep[a] + dep[b] - (min(mnd[t][l], mnd[t][r-(1<<t)+1]) << 1);
}

#define maxnode 1800010

struct Node {
	int v, r, siz;
	Node() {}
	Node(int _, int __): v(_), r(__) {}
};
struct Treap {
	Node ns[maxnode];
	int ToT, fa[maxnode], ch[maxnode][2];

	void maintain(int o) {
		if(!o) return ;
		ns[o].siz = 1;
		for(int i = 0; i < 2; i++) if(ch[o][i])
			ns[o].siz += ns[ch[o][i]].siz;
		return ;
	}
	void rotate(int u) {
		int y = fa[u], z = fa[y], l = 0, r = 1;
		if(z) ch[z][ch[z][1]==y] = u;
		if(ch[y][1] == u) swap(l, r);
		fa[u] = z; fa[y] = u; fa[ch[u][r]] = y;
		ch[y][l] = ch[u][r]; ch[u][r] = y;
		maintain(y); maintain(u);
		return ;
	}
	void Insert(int& o, int v) {
		if(!o) {
			ns[o = ++ToT] = Node(v, rand());
			return maintain(o);
		}
		bool d = v > ns[o].v;
		Insert(ch[o][d], v); fa[ch[o][d]] = o;
		if(ns[ch[o][d]].r > ns[o].r) {
			int t = ch[o][d];
			rotate(t); o = t;
		}
		return maintain(o);
	}
	int query(int o, int v) {
		if(!o) return 0;
		int ls = ch[o][0] ? ns[ch[o][0]].siz : 0;
		if(v < ns[o].v) return query(ch[o][0], v);
		return ls + 1 + query(ch[o][1], v);
	}
} sol;

int f[maxn], siz[maxn], rt, size;
bool vis[maxn];
void getroot(int u, int pa) {
	siz[u] = 1; f[u] = 0;
	for(int e = head[u]; e; e = nxt[e]) if(to[e] != pa && !vis[to[e]]) {
		getroot(to[e], u);
		siz[u] += siz[to[e]];
		f[u] = max(f[u], siz[to[e]]);
	}
	f[u] = max(f[u], size - siz[u]);
	if(f[rt] > f[u]) rt = u;
	return ;
}
void dfs(int u, int pa) {
	siz[u] = 1;
	for(int e = head[u]; e; e = nxt[e]) if(to[e] != pa && !vis[to[e]])
		dfs(to[e], u), siz[u] += siz[to[e]];
	return ;
}
int fa[maxn];
void solve(int u) {
	vis[u] = 1;
	for(int e = head[u]; e; e = nxt[e]) if(!vis[to[e]]) {
		dfs(to[e], u);
		f[rt = 0] = size = siz[to[e]]; getroot(to[e], u);
		fa[rt] = u; solve(rt);
	}
	return ;
}

int Rt[maxn], Rtfa[maxn];
void update(int s) {
	sol.Insert(Rt[s], 0);
	for(int u = s; fa[u]; u = fa[u]) {
		int d = cdist(s, fa[u]);
//		printf("insert:: %d(fa: %d): %d\n", u, fa[u], d);
		sol.Insert(Rtfa[u], d);
		sol.Insert(Rt[fa[u]], d);
	}
	return ;
}
int query(int s, int x) {
	int ans = sol.query(Rt[s], x);
//	printf("base: %d\n", ans);
	for(int u = s; fa[u]; u = fa[u]) {
		int d = cdist(s, fa[u]);
//		printf("query:: %d(fa: %d): %d (%d - %d)\n", u, fa[u], x - d, sol.query(Rt[fa[u]], x - d), sol.query(Rtfa[u], x - d));
		ans += sol.query(Rt[fa[u]], x - d) - sol.query(Rtfa[u], x - d);
	}
	return ans;
}

int main() {
	n = read(); int K = read(), sum = 0;
	for(int i = 1; i < n; i++) {
		int a = read(), b = read(), c = read();
		AddEdge(a, b, c); sum += c;
	}

	build(1, 0);
	rmq_init();
	f[rt = 0] = size = n; getroot(1, 0);
	solve(rt);
//	for(int i = 1; i <= n; i++) printf("(%d)%d%c", i, fa[i], i < n ? ‘ ‘ : ‘\n‘);
	for(int i = 1; i <= n; i++) update(i);
	for(int i = 1; i <= n; i++) {
		int l = 1, r = sum;
		while(l < r) {
			int mid = l + r >> 1;
			if(query(i, mid) - 1 < K) l = mid + 1; else r = mid;
		}
		printf("%d\n", l);
	}

	return 0;
}

又 1A 啦 2333333！