【UOJ #150】【NOIP 2015】运输计划

http://uoj.ac/problem/150

用树链剖分求lca，二分答案树上差分判断。

时间复杂度$O(nlogn)$，n，m同阶。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 300003;
int in() {
	int k = 0, fh = 1; char c = getchar();
	for(; c < '0' || c > '9'; c = getchar())
		if (c == '-') fh = -1;
	for(; c >= '0' && c <= '9'; c = getchar())
		k = (k << 3) + (k << 1) + c - '0';
	return k * fh;
}

struct QQ {int u, v, lca, len;} Q[N];
struct node {int nxt, to, w;} E[N << 1];
int n, m, cnt = 0, point[N], size[N], son[N], top[N], deep[N], fa[N];
int DFN[N], tot = 0, dis[N], fa_dis[N];

void ins(int u, int v, int w) {
	E[++cnt] = (node) {point[u], v, w}; point[u] = cnt;
}

void _(int x) {
	size[x] = 1; DFN[++tot] = x;
	for(int i = point[x]; i; i = E[i].nxt)
		if (E[i].to != fa[x]) {
			fa[E[i].to] = x;
			deep[E[i].to] = deep[x] + 1;
			dis[E[i].to] = dis[x] + E[i].w;
			fa_dis[E[i].to] = E[i].w;
			_(E[i].to);
			size[x] += size[E[i].to];
			if (size[E[i].to] > size[son[x]])
				son[x] = E[i].to;
		}
}

void __(int x) {
	if (!son[x]) return;
	top[son[x]] = top[x];
	__(son[x]);
	for(int i = point[x]; i; i = E[i].nxt)
		if (E[i].to != son[x] && E[i].to != fa[x])
			{top[E[i].to] = E[i].to; __(E[i].to);}
}

int LCA(int u, int v) {
	while (top[u] != top[v]) {
		if (deep[top[u]] < deep[top[v]]) swap(u, v);
		u = fa[top[u]];
	}
	return deep[u] < deep[v] ? u : v;
}

int del[N], cont, maxnow;

bool check(int s) {
	cont = 0; maxnow = 0;
	memset(del, 0, sizeof(int) * (n + 1));
	for(int i = 1; i <= m; ++i)
		if (Q[i].len > s) {
			++del[Q[i].u];
			++del[Q[i].v];
			del[Q[i].lca] -= 2;
			++cont;
			maxnow = max(maxnow, Q[i].len - s);
		}
	
	if (!cont) return true;
	for(int i = n; i > 1; --i) del[fa[DFN[i]]] += del[DFN[i]];
	for(int i = 2; i <= n; ++i)
		if (fa_dis[i] >= maxnow && del[i] == cont)
			return true;
	return false;
}

int main() {
	n = in(); m = in();
	int u, v, w;
	for(int i = 1; i < n; ++i) {
		u = in(); v = in(); w = in();
		ins(u, v, w);
		ins(v, u, w);
	}
	
	_(1);
	top[1] = 1; __(1);
	
	int left = 0, right = 0, mid;
	for(int i = 1; i <= m; ++i) {
		Q[i].u = in(); Q[i].v = in();
		Q[i].lca = LCA(Q[i].u, Q[i].v);
		Q[i].len = dis[Q[i].u] + dis[Q[i].v] - (dis[Q[i].lca] << 1);
		right = max(right, Q[i].len);
	}
	
	while (left < right) {
		mid = (left + right) >> 1;
		if (check(mid)) right = mid;
		else left = mid + 1;
	}
	
	printf("%d\n", left);
	return 0;
}

QwQ