树上形态改变统计贡献：1025T4

http://cplusoj.com/d/senior/p/SS231025D

答案为 $\sum w[x]-w[son[x]]$，$x$ 非儿子

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要维护断边，LCT固然可以，但不一定需要

发现如果发生了变化，只会由重儿子变成次重儿子

所以我们首先要维护次重儿子

同时我们拿树状数组维护其所有祖先的重儿子与次重儿子之差。

此时我们只需要在树状数组对应位置进行查询即可

#include
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar(); while(ch<'0'||
ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
//mt19937 rand(time(0));
//mt19937_64 rand(time(0));
//srand(time(0));
#define N 500010
//#define M
//#define mo
struct node {
	int y, id; 
};
int n, m, i, j, k, T;
int ans[N], w[N], nxt[N], totans, son[N], sum[N]; 
int u, v, su, sv, flg; 
vector<node>G[N]; 

struct Binary_tree {
	int cnt[N], sex; 
	void add(int x, int y) {
//		if(sex) printf("Add %d : %d\n", x, y); 
		if(!x) {
			cnt[0]+=y; return; 
		}
		while(x<=n) cnt[x]+=y, x+=x&-x; 
	}
	int que(int x) {
		int ans=0; 
		while(x) ans+=cnt[x], x-=x&-x; 
		return ans+cnt[0]; 
	}
}Bin, B1;

void dfs1(int x, int fa) {
	w[x]=1;
	for(auto t : G[x]) {
		int y = t.y; 
		if(y == fa) continue; 
		dfs1(y, x); w[x]+=w[y]; 
		sum[x]+=sum[y]; 
		if(w[y]>w[son[x]]) nxt[x]=son[x], son[x]=y; 
		else if(w[y]>w[nxt[x]]) nxt[x]=y; 
	}
	if(son[x]) totans+=w[x]-w[son[x]], sum[x]+=w[x]-w[son[x]]; 
//	printf("sum[%lld] = %lld || %lld %lld || %d\n", x, sum[x], son[x], nxt[x], w[x]);
}

void dfs2(int x, int fa, int dep, int p) {
	for(auto t : G[x]) {
		int y = t.y, id = t.id; 
		if(y == fa) continue; 
//		printf("%d->%d\n", x, y); 
//		dfs2(y, x); 
		if(y == son[x]) {
			Bin.add(w[son[x]]-w[nxt[x]], w[son[x]]-w[nxt[x]]); 
			B1.add(w[son[x]]-w[nxt[x]], 1); 
			ans[id]=totans-sum[y]+Bin.que(w[y])-w[y]*dep+(p+1-B1.que(w[y]))*w[y]; 
			if(!nxt[x]) --ans[id]; 
//			if(id==4) printf("%d(totans) %d(-sum[y]) %d(_change_son) %d(-size) %d(+son)\n", 
//				totans, sum[y], Bin.que(w[y]), w[y]*dep, (p+1-B1.que(w[y]))*w[y]); 
			dfs2(y, x, dep+1, p+1); 

			Bin.add(w[son[x]]-w[nxt[x]], -(w[son[x]]-w[nxt[x]])); 
			B1.add(w[son[x]]-w[nxt[x]], -1); 
		}
		else {
//			if(id==2) printf("%d(totans) %d(-sum[y]) %d(_change_son) %d(-size) %d(+son)\n", 
//				totans, sum[y], Bin.que(w[y]), w[y]*dep, (p-B1.que(w[y]))*w[y]); 
			ans[id]=totans-sum[y]+Bin.que(w[y])-w[y]*dep+(p-B1.que(w[y]))*w[y]; 
//			B1.add(w[son[x]]-w[nxt[x]], 1); 
			dfs2(y, x, dep+1, p); 
//			B1.add(w[son[x]]-w[nxt[x]], -1); 
		}
	}
}

signed main()
{
//	freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
		freopen("tree.in", "r", stdin);
	freopen("tree.out", "w", stdout);
//	T=read();
//	while(T--) {
//
//	}
	n=read(); 
	for(i=1; i<n; ++i) {
		u=read(); v=read(); 
		if(i==1) su=u, sv=v; 
		G[u].pb({v, i}); G[v].pb({u, i}); 
	}
	Bin.sex=1; 
	dfs1(su, sv); dfs1(sv, su); 
	totans+=n-max(w[su], w[sv])-1; 
//	printf("> %d\n", totans); 
	if(w[su]>w[sv]) {
		Bin.add(w[su]-w[sv], w[su]-w[sv]); 
		B1.add(w[su]-w[sv], 1); 
		flg=1; 
	}
	dfs2(su, sv, 2, flg);
	if(w[su]>w[sv]) {
		Bin.add(w[su]-w[sv], -(w[su]-w[sv])); 
		B1.add(w[su]-w[sv], -1); 
		flg=0; 
	}
	
//	printf("# %lld\n", sv); 
	if(w[sv]>w[su]) {
		Bin.add(w[sv]-w[su], (w[sv]-w[su])); 
		B1.add(w[sv]-w[su], 1); 
		flg=1; 
	}
	dfs2(sv, su, 2, flg);
	if(w[sv]>w[su]) {
		Bin.add(w[sv]-w[su], -(w[sv]-w[su])); 
		B1.add(w[sv]-w[su], -1); 
		flg=0; 
	}
	for(i=2; i<n; ++i) printf("%d\n", ans[i]); 
	return 0;
}