Codeforces Gym - 101147J Whistle's New Car

Discription

Statements

Whistle has bought a new car, which has an infinite fuel tank capacity.

He discovered an irregular country since it has n cities and there are exactly n - 1roads between them, of course, all cities are connected. He is so much clever, he realized that the country is like a rooted tree of n nodes and node 1 is the root. Each city i has only one filling station by which he can fill his car's fuel tank in no more than X_i liter. Whistle liked the country very much, and he wants to know what the most attractive city in the country is. The attractiveness of the city i is defined by how much it’s reachable from other cities, in other words the attractiveness of city is the number of cities j that satisfies these condition:

• City j is in the subtree of city i (except for city i itself).
• Whistle will start at city j and will only fill his car’s fuel tank with X_jliters and never fill it again until he reach city i.
• Whistle should be able to reach city i with non-negative fuel.

He knows the length of every road and that 1 Km will take exactly 1 liter on any road.

As you know, Whistle is very clever, but he is not that good at programming, so he asked you to help him. He wants to know the attractiveness of each city, so that he can decide which city to live in.

Input

The first line of input contains one integer T, the number of test cases.

The next line contains one integer (1 ≤ n ≤ 500, 000), The number of cities in the country.

The next line contains n integers (1 ≤ X_i ≤ 1, 000, 000, 000).

Each one of the next n - 1 line contains three integers A, B, C (1 ≤ A, B ≤ n and 1 ≤ C ≤ 1, 000, 000, 000), that means there is a road between city A and city B of length C.

Output

For each test case, output a line containing n integers, the attractiveness of each city.

Example

Input

1
4
5 10 5 10
1 2 100
2 3 5
3 4 5

Output

0 2 1 0

Note

Large I/O files. Please consider using fast input/output methods.

 

    (为什么是文件输入标准输出hhhh，被坑了好久。。。)

    每个点维护一个小根堆，往树上父亲合并的时候要先把这个堆都打个  -val_to_fa 的标记。因为涉及到合并和打标机，所以我们写一下左偏树就好啦。

    最后每个点的答案就是 这个点的左偏树大小-1。

 

    这个题还有树剖做法，，，虽然很好想(直接考虑每个点向上的影响就行了)，但是因为跑的太慢而被我的可并堆艹爆hhhhh

于是我就成了GYM上第二快的人了hhhh(第一是个丧病各种写宏的毒瘤人士hhh)

 

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=500005;
int to[maxn*2],ne[maxn*2],val[maxn*2],num;
int siz[maxn],hd[maxn],L[maxn],ans[maxn];
int n,T,f[maxn],lc[maxn],rc[maxn];
ll W[maxn],tag[maxn];
inline void add(int x,int y,int z){ to[++num]=y,ne[num]=hd[x],hd[x]=num,val[num]=z;}
inline int read(){
	int x=0; char ch=getchar();
	for(;!isdigit(ch);ch=getchar());
	for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x;
}
void Wt(int x){ if(x>=10) Wt(x/10); putchar(x%10+'0');}
inline void init(){ num=0,memset(hd,0,sizeof(hd));}
inline void update(int x,ll y){ tag[x]+=y,W[x]+=y;}
inline void pushdown(int x){
    if(tag[x]){
	    if(lc[x]) update(lc[x],tag[x]);
	    if(rc[x]) update(rc[x],tag[x]); 
		tag[x]=0;
	}
}

int merge(int x,int y){
	if(!x||!y) return x+y;
	pushdown(x),pushdown(y);
	
	if(W[x]>W[y]) swap(x,y);
	rc[x]=merge(rc[x],y),f[rc[x]]=x;
	if(L[rc[x]]>L[lc[x]]) swap(lc[x],rc[x]);
	L[x]=L[rc[x]]+1,siz[x]=siz[lc[x]]+siz[rc[x]]+1;
	
	return x;
}

int DEL(int x){
	pushdown(x),f[lc[x]]=f[rc[x]]=0;
	return merge(lc[x],rc[x]);
}

int dfs(int x,int fa){
	int root=x,TO;
	for(int i=hd[x];i;i=ne[i]) if(to[i]!=fa){
		TO=dfs(to[i],x),update(TO,-val[i]);
		while(W[TO]<0) TO=DEL(TO);
		root=merge(root,TO);
	}
	
	ans[x]=siz[root]-1;
	return root;
}

int main(){
	freopen("car.in","r",stdin);
//	freopen("data.out","w",stdout);
	scanf("%d",&T);
	while(T--){
		int uu,vv,ww;
		init(),scanf("%d",&n);
		for(int i=1;i<=n;i++) W[i]=read(),tag[i]=lc[i]=rc[i]=L[i]=f[i]=0,siz[i]=1;
		for(int i=1;i<n;i++) uu=read(),vv=read(),ww=read(),add(uu,vv,ww),add(vv,uu,ww);
		dfs(1,-1);
		for(int i=1;i<=n;i++) Wt(ans[i]),putchar(' ');
		puts("");
	}
	return 0;
}