CF796C Bank Hacking 题解

洛谷链接

题目

Although Inzane successfully found his beloved bone, Zane, his owner, has yet to return. To search for Zane, he would need a lot of money, of which he sadly has none. To deal with the problem, he has decided to hack the banks.

There are n banks, numbered from 1 to n. There are also n - 1 wires connecting the banks. All banks are initially online. Each bank also has its initial strength: bank i has initial strength a i.

Let us define some keywords before we proceed. Bank i and bank j are neighboring if and only if there exists a wire directly connecting them. Bank i and bank j are semi-neighboring if and only if there exists an online bank k such that bank i and bank k are neighboring and bank k and bank j are neighboring.

When a bank is hacked, it becomes offline (and no longer online), and other banks that are neighboring or semi-neighboring to it have their strengths increased by 1.

To start his plan, Inzane will choose a bank to hack first. Indeed, the strength of such bank must not exceed the strength of his computer. After this, he will repeatedly choose some bank to hack next until all the banks are hacked, but he can continue to hack bank x if and only if all these conditions are met:

1. Bank x is online. That is, bank x is not hacked yet.
2. Bank x is neighboring to some offline bank.
3. The strength of bank x is less than or equal to the strength of Inzane's computer.

Determine the minimum strength of the computer Inzane needs to hack all the banks.

给定一棵带点权树，选出一个最佳的根节点，使得根节点的点权不变，它的儿子点权加1，其余点点权加2，并使最大点权最小，输出这个最小的最大点权

输入格式

The first line contains one integer \(n (1 ≤ n ≤ 3·10^5)\) — the total number of banks.

The second line contains \(n\) integers \(a_1, a_2, ..., a_n ( - 10^9 ≤ a i ≤ 10^9)\) — the strengths of the banks.

Each of the next n - 1 lines contains two integers \(u_i\) and \(v_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i)\) — meaning that there is a wire directly connecting banks \(u_i\) and \(v_i\).

输出格式

Print one integer — the minimum strength of the computer Inzane needs to accomplish the goal.

It is guaranteed that the wires connect the banks in such a way that Inzane can somehow hack all the banks using a computer with appropriate strength.

题解

给定一棵带点权树，选出一个最佳的根节点，使得根节点的点权不变，它的儿子点权加1，其余点点权加2，并使最大点权最小，输出这个最小的最大点权

这是洛谷上的翻译, 很简洁明了.

这道题直接遍历一遍树, 按照题意统计一下就好了.

输入的时候先求出点权的最大值和每个权值的个数, 然后遍历每一个节点, 每个节点遍历每一个子节点, 根据题目性质计算即可.

代码

#include <bits/stdc++.h>
using namespace std;
struct Edge { int to, next; } edges[600010];
int n, tot, ans, maxv = -0x3f3f3f3f, flag, x, y, a[300010], head[300010];
map<int, int> m;
void add(int x,int y) { edges[++tot] = (Edge){y, head[x]},head[x] = tot; }
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]), m[a[i]]++, maxv = max(maxv, a[i]);
    ans = maxv + 2;
    for (int i = 1; i < n; i++) scanf("%d%d", &x, &y), add(x,y), add(y,x);
    for (int i = 1; i <= n; i++) {
        flag = 0;
        for (int j = head[i]; j; j = edges[j].next) {
            m[a[edges[j].to]]--;
            if (a[edges[j].to] == maxv) flag = 1;
        }
        if (!m[maxv]) ans = maxv + 1;
        if (maxv == a[i] && m[maxv] == 1) {
            if (m[maxv - 1] || flag) // 其它节点有ans-1的权值,-1+2=+1 或者 子节点中也有最大值,子节点+1后是答案
                ans = maxv + 1;
            else {
                ans = maxv; // 根节点是最大值, 并且子树中没有更大的, 直接break
                break;
            }
        }
        for (int j = head[i]; j; j = edges[j].next) m[a[edges[j].to]]++;
    }
    printf("%d\n", ans);
    return 0;
}