题意：太复杂···
思路：每个银行最多只能加两次，起点一次都不加，起点相邻的点只加一次，剩下的加两次，所以只需要二分法最小化最大值。。。坑点有点多，代码自行体会···
#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
const int maxn = 300000+10;
const long long INF = 1e10+1;
int n;
long long arr[maxn];
vector<int>vis[maxn];
bool check(long long x)
{
int tmp = 0;
for(int i=1;i<=n;i++)
{
if(arr[i]>x)
return false;
if(arr[i]+2>x)
tmp++;
}
for(int i = 1;i<=n;i++)
{
int ans = tmp;
if(arr[i]+2>x)
ans--;
for(int j = 0;j<vis[i].size();j++)
{
int b = vis[i][j];
if(arr[b]+1==x)
ans--;
}
if(ans==0)
return true;
}
return false;
}
int main()
{
scanf("%d",&n);
long long maxx = -INF;
long long minn = INF;
for(int i = 1;i<=n;i++)
{
scanf("%lld",&arr[i]);
maxx = max(arr[i],maxx);
minn = min(arr[i],minn);
}
for(int i = 1;i<=n-1;i++)
{
int a,b;
scanf("%d%d",&a,&b);
vis[a].push_back(b);
vis[b].push_back(a);
}
long long l = minn,r = maxx+2;
long long gg = l;
while(l<=r)
{
long long mid = (l+r)/2;
if(check(mid))
{
gg = mid;
r = mid-1;
}
else
l = mid+1;
}
printf("%lld",gg);
return 0;
}
