meeting

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题目链接

题意

有一颗$n$个点的树，其中有$k$个点有人居住，树上每条边花费都是1，现要找到一个点，使得所有人到这个点的花费的最大值最小。

思路

树形dp

每个点作为答案的话，我们可以先$dfs$一遍算出，所有点的子树中到有人的点到它的最远距离。
接下来考虑，第二遍$dfs$，我们要计算，来自他的父亲的有人的点到他的最远距离。
第二遍dp比较复杂，需要考虑较多东西，要仔细一点。

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#ifndef ONLINE_JUDGE
#define dbg(x...) do{cout << "\033[33;1m" << #x << "->" ; err(x);} while (0)
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
#else
#define dbg(...)
#endif
#define inf 1ll << 50
 
const int N = 1e5 + 5;
int tot;
int head[N];
struct E
{
    int v, nxt, mx;
}edge[2 * N];
int son[N];
bool is[N];
int mx[N];
 
void add_edge(int u, int v)
{
    edge[++tot].v = v;
    edge[tot].nxt = head[u];
    head[u] = tot;
}
 
void dfs(int u, int f)
{
    for (int it = head[u]; it != -1; it = edge[it].nxt)
    {
        int v = edge[it].v;
        if (v == f)
            continue;
        dfs(v, u);
        if (mx[v] != -1 && (son[u] == -1 || mx[u] < mx[v] + 1))
        {
            son[u] = v;
            mx[u] = mx[v] + 1;
        }
        if (is[v] && (son[u] == -1 || mx[u] < 1))
        {
            son[u] = v;
            mx[u] = 1;
        }
    }
    //dbg(u, mx[u]);
}
int ans;
 
void dfs2(int u, int f, int t)
{
    int pmx = -1, qmax = max(mx[u], t);
    if (qmax == -1)
        qmax = -1;
    else
        qmax += 1;
    if (is[u])
        pmx = max(pmx, 0), qmax = max(qmax, 1);
    for (int it = head[u]; it != -1; it = edge[it].nxt)
    {
        int v = edge[it].v;
        if (v == f)
            continue;
        if (v != son[u])
        {
            if (mx[v] != -1)
                pmx = max(pmx, mx[v] + 1);
            if (is[v])
                pmx = max(pmx, 1);
            dfs2(v, u, qmax);
        }
    }
    //dbg(u, pmx, qmax, t, max(t, pmx));
    if (son[u] != -1)
        dfs2(son[u], u, max(t, pmx) == -1? -1 : max(t, pmx) + 1);
    mx[u] = max(mx[u], t);
    if (mx[u] < mx[ans])
        ans = u;
    //dbg(u, mx[u], ans);
}
int main()
{
    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++)
        head[i] = -1, son[i] = -1, mx[i] = -1;
    for (int i = 1; i < n; i++)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        add_edge(u, v);
        add_edge(v, u);
    }
    for (int i = 1; i <= k; i++)
    {
        int x;
        scanf("%d", &x);
        is[x] = 1;
    }
    if (k == 0)
    {
        puts("0");
        return 0;
    }
    dfs(1, 0);
    ans = 1;
    dfs2(1, 0, -1);
    printf("%d\n", mx[ans]);
    return 0;
}

直径

也可以在这些点中找直径，若直径是$d$，那么这些点都可以在内到达这点。
这种方法比较好写。

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#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int mx[N];
int is[N];
struct E
{
	int v, nxt;
}edge[2 * N];
int tot;
int head[N];
void add_edge(int u, int v)
{
	edge[++tot].v = v;
	edge[tot].nxt = head[u];
	head[u] = tot;
}
int dep[N];
void dfs(int u, int f)
{
	mx[u] = 0;
	dep[u] = dep[f] + 1;
	for (int it = head[u]; it != -1; it = edge[it].nxt)
	{
		int v = edge[it].v;
		if (v == f)
			continue;
		dfs(v, u);
		if (is[v])
			if (mx[u] == 0)
				mx[u] = v;
		if (mx[v])
			mx[u] = dep[mx[u]] - dep[u] >= dep[mx[v]] - dep[u] + 1? mx[u] : mx[v];
	}
}

int main()
{
	int n, k;
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; i++)
	{
		head[i] = -1;
	}
	for (int i = 1; i < n; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		add_edge(u, v);
		add_edge(v, u);
	}
	for (int i = 1; i <= k; i++)
	{
		int x;
		scanf("%d", &x);
		is[x] = 1;
	}
	dep[0] = 0;
	dfs(1, 0);
	int rt = mx[1];
	dfs(rt, 0);
	printf("%d\n", (dep[mx[rt]] - dep[rt] + 1) / 2);
	return 0;
}

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