【LG1600】[NOIP2016]天天爱跑步

题面

洛谷

题解

考虑一条路径\(S\rightarrow T\)是如何给一个观测点\(x\)造成贡献的，

一种是从\(x\)的子树内出来，另外一种是从\(x\)的子树外进去。

令\(S,T\)的最近公共祖先为\(lca\)，那么这条路径可表示为\(S\rightarrow lca\rightarrow T\)（如果\(lca=S\;or\;T\)可以特判）。

考虑两种情况如何贡献，

首先在\(S\rightarrow lca\)上的点，需要满足\(dep_S-dep_x=w_x\)，

而对于\(lca\rightarrow T\)上的点，需要满足\((dep_S-dep_{lca})+(dep_x-dep_{lca})=w_x\Leftrightarrow dep_S-2dep_{lca}=w_x-dep_x\)。

这样的话，对于一条路径，我们可以拆成两条分别对其进行差分，在用一颗线段树在其对应位置上\(\pm 1\)，然后线段树合并在对应位置上查即可。

具体实现细节详见代码。

代码

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <vector>

using namespace std;

inline int gi() {

    register int data = 0, w = 1;

    register char ch = 0;

    while (!isdigit(ch) && ch != '-') ch = getchar();

    if (ch == '-') w = -1, ch = getchar();

    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();

    return w * data;

}

const int MAX_N = 3e5 + 5;

struct Graph { int to, next; } e[MAX_N << 1];

int fir[MAX_N], e_cnt;

void clearGraph() { memset(fir, -1, sizeof(fir)); e_cnt = 0; }

void Add_Edge(int u, int v) { e[e_cnt] = (Graph){v, fir[u]}, fir[u] = e_cnt++; }

int fa[MAX_N], top[MAX_N], dep[MAX_N], size[MAX_N], son[MAX_N];

void dfs1(int x) {

	size[x] = 1, dep[x] = dep[fa[x]] + 1;

	for (int i = fir[x]; ~i; i = e[i].next) {

		int v = e[i].to; if (v == fa[x]) continue;

		fa[v] = x, dfs1(v), size[x] += size[v];

		if (size[son[x]] < size[v]) son[x] = v;

	}

}

void dfs2(int x, int tp) {

	top[x] = tp;

	if (son[x]) dfs2(son[x], tp);

	for (int i = fir[x]; ~i; i = e[i].next) {

		int v = e[i].to; if (v == son[x] || v == fa[x]) continue;

		dfs2(v, v);

	}

}

int LCA(int x, int y) {

	while (top[x] != top[y]) {

		if (dep[top[x]] < dep[top[y]]) swap(x, y);

		x = fa[top[x]];

	}

	return dep[x] < dep[y] ? x : y;

}

struct Path { int s, t, lca; } p[MAX_N];

int N, M, w[MAX_N], ans[MAX_N];

vector<int> Add1[MAX_N], Del1[MAX_N], Add2[MAX_N], Del2[MAX_N];

struct Node { int ls, rs, v; } t[MAX_N << 6];

int rt1[MAX_N], rt2[MAX_N], tot;

void insert(int &o, int l, int r, int pos, int op) {

	if (!o) o = ++tot;

	t[o].v += op;

	if (l == r) return ;

	int mid = (l + r) >> 1;

	if (pos <= mid) insert(t[o].ls, l, mid, pos, op);

	else insert(t[o].rs, mid + 1, r, pos, op);

}

int merge(int x, int y, int l, int r) {

	if (!x || !y) return x | y;

	if (l == r) return t[x].v += t[y].v, x;

	int mid = (l + r) >> 1;

	t[x].ls = merge(t[x].ls, t[y].ls, l, mid);

	t[x].rs = merge(t[x].rs, t[y].rs, mid + 1, r);

	return t[x].v = t[t[x].ls].v + t[t[x].rs].v, x;

}

int query(int o, int l, int r, int pos) {

	if (!o) return 0;

	if (l == r) return t[o].v;

	int mid = (l + r) >> 1;

	if (pos <= mid) return query(t[o].ls, l, mid, pos);

	else return query(t[o].rs, mid + 1, r, pos);

}

void Dfs(int x) {

	for (int i = fir[x]; ~i; i = e[i].next) {

		int v = e[i].to; if (v == fa[x]) continue;

		Dfs(v);

		rt1[x] = merge(rt1[x], rt1[v], -N, N << 1);

		rt2[x] = merge(rt2[x], rt2[v], -N, N << 1);

	}

	for (int i : Add1[x]) insert(rt1[x], -N, N << 1, i, 1);

	for (int i : Add2[x]) insert(rt2[x], -N, N << 1, i, 1);

	for (int i : Del1[x]) insert(rt1[x], -N, N << 1, i, -1);

	for (int i : Del2[x]) insert(rt2[x], -N, N << 1, i, -1);

	ans[x] = query(rt1[x], -N, N << 1, w[x] + dep[x]) + query(rt2[x], -N, N << 1, w[x] - dep[x]);

}

int main () {

#ifndef ONLINE_JUDGE

    freopen("cpp.in", "r", stdin);

#endif

	clearGraph();

	N = gi(), M = gi();

	for (int i = 1; i < N; i++) {

		int u = gi(), v = gi();

		Add_Edge(u, v), Add_Edge(v, u);

	}

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

	for (int i = 1; i <= N; i++) w[i] = gi();

	for (int i = 1; i <= M; i++) {

		int s = gi(), t = gi(), lca = LCA(s, t);

		int d1 = dep[s], d2 = -dep[s];

		if (lca == t) { Add1[s].push_back(d1), Del1[fa[lca]].push_back(d1); continue; }

		if (lca == s) { Add2[t].push_back(d2), Del2[fa[lca]].push_back(d2); continue; }

		d2 = dep[s] - 2 * dep[lca];

		Add1[s].push_back(d1), Del1[fa[lca]].push_back(d1);

		Add2[t].push_back(d2), Del2[lca].push_back(d2);

	}

	Dfs(1);

	for (int i = 1; i <= N; i++) printf("%d ", ans[i]);

	putchar('\n');

    return 0;

}

巴特西

【LG1600】[NOIP2016]天天爱跑步

【LG1600】[NOIP2016]天天爱跑步

题面

题解

代码

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