ARC115F Migration

二分答案。

鉴于 $n$ 比较小，先 $\mathcal{O}(n^2)$ 预处理出每个点最优可跳到哪里，使得路径最大值尽量小，保证不能跳到原点且权值小于等于根且编号尽量小。

考虑二分答案 $lim$，每个分别从开始节点和终止节点跳最优点，跳到权值和超过 $lim$ 为止，最后判断对应的两点是否相等即可，单次时间复杂度 $\mathcal{O}(kn)$。

时间复杂度 $\mathcal{O}(n^2\log V)$。

#include<bits/stdc++.h>

using namespace std;

#define int long long
#define ha putchar(' ')
#define he putchar('\n')
typedef long long ll;

inline int read()
{
    
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
    
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = x * 10 + c - '0', c = getchar();
    return x * f;
}

inline void write(int x)
{
    
    if(x < 0)
    {
    
        putchar('-');
        x = -x;
    }
    if(x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}

const int _ = 4010;

int n, k, w[_], a[_], A[_], B[_], val[_], f[_], rt, tot, head[_], to[_ << 1], nxt[_ << 1], s[_], t[_];

void add(int u, int v)
{
    
	to[++tot] = v, nxt[tot] = head[u], head[u] = tot;
}

void dfs(int u, int fa, int z)
{
    
	z = max(z, w[u]), a[u] = z;
	if(w[u] < w[rt] || (w[u] == w[rt] && u < rt))
		if(!f[rt] || a[u] < a[f[rt]]) f[rt] = u;
	for(int i = head[u]; i; i = nxt[i])
		if(to[i] ^ fa) dfs(to[i], u, z);
}

bool check(int x)
{
    
	int s1 = 0, s2 = 0;
	for(int i = 1; i <= k; ++i) A[i] = s[i], B[i] = t[i], s1 += w[A[i]], s2 += w[B[i]];
	if(s1 > x || s2 > x) return 0;
	while (1)
	{
    
		bool flg = 0;
		for(int i = 1; i <= k; ++i)
			if (f[A[i]] && s1 - w[A[i]] + val[A[i]] <= x) s1 = s1 - w[A[i]] + w[f[A[i]]], A[i] = f[A[i]], flg = 1;
		for(int i = 1; i <= k; ++i)
			if (f[B[i]] && s2 - w[B[i]] + val[B[i]] <= x) s2 = s2 - w[B[i]] + w[f[B[i]]], B[i] = f[B[i]], flg = 1;
		if(!flg) break;
	}
	for(int i = 1; i <= k; ++i)
		if (A[i] ^ B[i]) return 0;
	return 1;
}

signed main()
{
    
	n = read();
	for(int i = 1; i <= n; ++i) w[i] = read();
	for(int i = 1, u, v; i < n; ++i)
	{
    
		u = read(), v = read();
		add(u, v), add(v, u);
	}
	k = read();
	for(int i = 1; i <= k; ++i) s[i] = read(), t[i] = read();
	for(int i = 1; i <= n; ++i)
	{
    
		rt = i;
		memset(a, 0, sizeof a);
		dfs(i, i, w[i]);
		val[i] = a[f[i]];
	}
	int l = 0, r = 2e12, mid;
	ll ans = 0;
	while(l <= r)
	{
    
		mid = (l + r) >> 1;
		if(check(mid)) ans = mid, r = mid - 1;
		else l = mid + 1;
	}
	write(ans), he;
	return 0;
}