H.Crystalfly

H. Crystalfly
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Paimon is catching crystalflies on a tree, which are a special kind of butterflies in Teyvat. A tree is a connected graph consisting of n vertices and (n−1) undirected edges.

Pixiv ID: 93964680

There are initially ai crystalflies on the i-th vertex. When Paimon reaches a vertex, she can catch all the remaining crystalflies on the vertex immediately. However, the crystalflies are timid. When Paimon reaches a vertex, all the crystalflies on the adjacent vertices will be disturbed. For the i-th vertex, if the crystalflies on the vertex are disturbed for the first time at the beginning of the t′-th second, they will disappear at the end of the (t′+ti)-th second.

At the beginning of the 0-th second, Paimon reaches vertex 1 and stays there before the beginning of the 1-st second. Then at the beginning of each following second, she can choose one of the two operations:

Move to one of the adjacent vertices of her current vertex and stay there before the beginning of the next second (if the crystalflies in the destination will disappear at the end of that second she can still catch them).
Stay still in her current vertex before the beginning of the next second.
Calculate the maximum number of crystalflies Paimon can catch in 1010101010 seconds.

Input
There are multiple test cases. The first line of the input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer n (1≤n≤105) indicating the number of vertices.

The second line contains n integers a1,a2,⋯,an (1≤ai≤109) where ai is the number of crystalflies on the i-th vertex.

The third line contains n integers t1,t2,⋯,tn (1≤ti≤3) where ti is the time before the crystalflies on the i-th vertex disappear after disturbed.

For the next (n−1) lines, the i-th line contains two integers ui and vi (1≤ui,vi≤n) indicating an edge connecting vertices ui and vi in the tree.

It’s guaranteed that the sum of n of all the test cases will not exceed 106.

Output
For each test case output one line containing one integer indicating the maximum number of crystalflies Paimon can catch.

Example
inputCopy
2
5
1 10 100 1000 10000
1 2 1 1 1
1 2
1 3
2 4
2 5
5
1 10 100 1000 10000
1 3 1 1 1
1 2
1 3
2 4
2 5
outputCopy
10101
10111
Note
For the first sample test case, follow the strategy below.

During the 0-th second
Paimon arrives at vertex 1;
Paimon catches 1 crystalfly;
Crystalflies in vertices 2 and 3 are disturbed.
During the 1-st second
Paimon arrives at vertex 3;
Paimon catches 100 crystalflies.
During the 2-nd second
Paimon arrives at vertex 1;
Crystalflies in vertex 2 disappears.
During the 3-rd second
Paimon arrives at vertex 2;
Crystalflies in vertices 4 and 5 are disturbed.
During the 4-th second
Paimon arrives at vertex 5;
Paimon catches 10000 crystalflies;
Crystalflies in vertex 4 disappears.
For the second sample test case, the optimal strategy is the same with the first sample test case. Crystalflies in vertex 2 are scheduled to disappear at the end of the 3-rd (instead of the 2-nd) second, allowing Paimon to catch them.

题目

• 给出一个n个点（1e5），以1位根的树。每个点有价值为a[i]的蝴蝶。
• 你从点1出发抓蝴蝶，每一秒可以移动到相邻的一个点。当你第一次进入某个点后，其相邻点的蝴蝶就会在t[v]（1-3）秒后消失。
• 问你最多能抓到多少价值的蝴蝶。

思路：

• 当我们初次到达某个节点 u 时，它的所有儿子会被激活，由于 t[i]<=3 ，所以我们的决策只有两种。
  一种是进入某个儿子 v 后继续向下，这样 x 的所有其它儿子的蝴蝶都无法被获取。
  另一种是进入某个儿子 v 获取 av 后立即回到 x，然后进入另一个儿子 w 并获得 aw ，这要求 tw=3。
• 我们令 f[u] 表示节点 u 在我们进入之前所有蝴蝶都已经消失，同时它的孩子在我们进入 u 之前都未被激活，这棵子树的最优答案，那我们要求的最终答案就是f[1]+a[1] 。
• 我们令 G[u] 表示 u 的儿子集合， sum[u] 表示 u 的所有儿子 v 的 f[v] 值之和。考虑两种决策对应的转移.
  第一种决策：f[u] = sum[u]+max{a[v]}, 从v走下去
  第二种决策：f[u] = sum[u]-f[v]+sum[v]+a[v]+max{a[v0], v0!=v, v==3}， 拿一个除了v最大的, 再从v走下去,v需要==3

#include
using namespace std;
typedef long long LL;
const int maxn = 2e5+10;
vectorG[maxn];
LL a[maxn], t[maxn];//f[x]:x的蝴蝶已经消失,儿子未被激活的最优答案. sum[x]=sum{f[y]}
LL f[maxn], sum[maxn]; 
void dfs(int x, int fa){multisetse;//1.进入y后一直向下, x所有儿子都不要了, 拿一个max{a[y]}, f[x] = sum[x]+mx{a[y]}LL mx = 0;for(int y : G[x]){if(y==fa)continue;dfs(y, x);sum[x] += f[y];mx = max(mx, a[y]);if(t[y]==3)se.insert(a[y]);}f[x] = sum[x] + mx;//2.对于t[y]==3,先拿一个其他的（除了他最大的），再从这里走下去se.insert(-1e18);for(int y: G[x]){if(y==fa)continue;if(t[y]==3)se.erase(se.find(a[y]));f[x] = max(f[x], sum[x]+*se.rbegin()-f[y]+sum[y]+a[y]);if(t[y]==3)se.insert(a[y]);}
}
int main(){ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int T;  cin>>T;while(T--){int n;  cin>>n;for(int i = 1; i <= n; i++)cin>>a[i];for(int i = 1; i <= n; i++){cin>>t[i];sum[i] = f[i] = 0;  G[i].clear();}for(int i = 1; i < n; i++){int u, v;  cin>>u>>v;G[u].push_back(v);G[v].push_back(u);}dfs(1, -1);cout<

#include
using namespace std;
typedef long long LL;
const LL N = 1e5+10;
LL a[N], n;
LL b[N], vis[N];
void add(LL x){while(x<=n){ b[x]++; x += x&(-x); } }
LL query(LL x){ LL res=0; while(x){ res+=b[x]; x-=x&(-x); } return res;}
int main(){ios::sync_with_stdio(0), cin.tie(0),cout.tie(0);LL T;  cin>>T;while(T--){cin>>n;for(LL i = 1; i <= n; i++){cin>>a[i];b[i] = vis[i] = 0;}cout<<0;vis[a[1]] = 1;  add(a[1]);LL ok = 0, cnt = 0, ans = 0;for(LL i = 2; i <= n; i++){if(!vis[a[i]])vis[a[i]] = 1, add(a[i]);if(a[i]==a[1])ok = 1;cnt += ok-(ok?a[i]>a[1]:0);if(a[i]>a[1]) ans += 1+cnt, swap(a[1],a[i]), cnt=ok=0;ans += query(a[1])-query(a[i]);cout<<" "<