Codeforces 796C 想法

Bank Hacking

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ai.

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 neighboringand 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 bankx 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.

Input

The first line contains one integer n (1 ≤ n ≤ 3·105) — the total number of banks.

The second line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the strengths of the banks.

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

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.

Output

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

Examples

input

51 2 3 4 51 22 33 44 5

output

5

input

738 -29 87 93 39 28 -551 22 53 22 41 77 6

output

93

input

51 2 7 6 71 55 33 42 4

output

8

Note

In the first sample, Inzane can hack all banks using a computer with strength 5. Here is how:

• Initially, strengths of the banks are [1, 2, 3, 4, 5].
• He hacks bank 5, then strengths of the banks become [1, 2, 4, 5,  - ].
• He hacks bank 4, then strengths of the banks become [1, 3, 5,  - ,  - ].
• He hacks bank 3, then strengths of the banks become [2, 4,  - ,  - ,  - ].
• He hacks bank 2, then strengths of the banks become [3,  - ,  - ,  - ,  - ].
• He completes his goal by hacking bank 1.

In the second sample, Inzane can hack banks 4, 2, 3, 1, 5, 7, and 6, in this order. This way, he can hack all banks using a computer with strength 93.

题意：给你一棵树  两家银行称为邻居  他们的距离是1

两家银行称为间接邻居  他们的距离是2且中点银行没有被hack

每次hack完一家银行后  他的邻居和间接邻居的防御值都+1

开始你可以任意选一家银行hack   接下来你只能hack已经hack过的银行的邻居

求最小攻击值能hack掉所有银行

题解：

这道题是细节题

令mix1为最大的防御值  mix2为次大的防御值

答案很明显  只有三种可能  mix1  mix1+1  mix1+2

当只有一家银行时  答案就是mix1    return 0

当只有两家银行时  答案就是max( mix1 ,mix2+1 )    return 0

当所有的点防御值都一样时  当银行呈烟花状时(一个银行是其他所有银行的邻居)  那么我们先hack这家银行  答案就是 mix1+1  return 0

          否则  答案就是mix1+2   return 0

当最大的防御值的数大于等于3个时  如果呈烟花状(一个烟花包含所有最大的防御值的银行)  答案就是mix1+1

否则  答案就是mix1+2

当最大的防御值的数等于2个时  如果这两个点距离小于等于2  那么答案就是mix1+1

否则  答案就是mix1+2

当最大的防御值的数等于1时  如果以这个点为中心和所有次大的防御值的银行呈烟花状  答案就是  mix1

否则  答案就是  max( mix1 , mix2+2 )

还有一种暴力做法(我听了想打人)

把这些点都放进multiset里面  然后for每个点  对于每个点  他相邻的点都+1  然后把这个点和相邻的都删掉  再从multiset中找出最大的+2  再把删去的点放回去

这样的做法复杂度是nlongn  因为每个点被放进去的次数是这个点的du+1

这里放分类讨论的代码

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<deque>using namespace std;struct node{int to,nex;}edge[600005];int head[300005],cnt,num[300005],du[300005];void add(int u,int v){edge[cnt].to=v;edge[cnt].nex=head[u];head[u]=cnt++;}deque<int>sp;int vis[300005];int main(){int n,i,j,x,y,mix1=-1000000001,mix2=-1000000001;scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&num[i]);if(num[i]>=mix1){mix2=mix1;mix1=num[i];}else if(num[i]>=mix2)mix2=num[i];}int num1=0,num2=0;for(i=1;i<=n;i++){if(mix1==num[i])num1++;else if(mix2==num[i])num2++;}int td=0,lab=0,mis=0;;memset(head,-1,sizeof(head));for(i=1;i<n;i++){scanf("%d%d",&x,&y);add(x,y);add(y,x);if(num[x]==mix1){            du[y]++;            if(du[y]>mis){                mis=du[y];                lab=y;            }}if(num[y]==mix1){            du[x]++;            if(du[x]>mis){                mis=du[x];                lab=x;            }}td=max(td,du[x]);td=max(td,du[y]);}if(num1==n){if(n==1)printf("%d\n",mix1);else if(td==n-1)printf("%d\n",mix1+1);else printf("%d\n",mix1+2);return 0;}if(n==1){printf("%d\n",mix1);return 0;}else if(n==2){if(num[1]==num[2])printf("%d\n",num[1]+1);else printf("%d\n",mix1);return 0;}if(num1>=3){        if(mis==num1||(mis==num1-1&&num[lab]==mix1))printf("%d\n",mix1+1);else printf("%d\n",mix1+2);}else if(num1==2){for(i=1;i<=n;i++){if(num[i]==mix1)break;}int flag=1;vis[i]=1;sp.push_back(i);while(!sp.empty()){            int f=sp.front();            sp.pop_front();            if(vis[f]>=4)continue;            if(num[f]==mix1&&f!=i){                flag=0;                break;            }            for(j=head[f];~j;j=edge[j].nex){                int v=edge[j].to;                if(vis[v])continue;                vis[v]=vis[f]+1;                sp.push_back(v);            }}if(flag){printf("%d\n",mix1+2);}else printf("%d\n",mix1+1);}else{int nums=0;for(i=1;i<=n;i++){if(num[i]==mix1)break;}for(j=head[i];~j;j=edge[j].nex){if(num[edge[j].to]==mix2)nums++;}if(nums==num2)printf("%d\n",mix1);else printf("%d\n",max(mix1,mix2+2));}return 0;}