题目链接：传送门

Description

给定一棵nn个点的树，对于每一个点求出子树内所在深度节点个数最多的相对深度（即深度减去当前点的深度）

Constraints

n≤106n \leq 10^6

Solution

用map记录子树信息，直接树上启发式合并即可

Debug

53L: 不能直接swap(ans[x], ans[y])，否则会改变ans[y]的值

58L: 不能用迭代器，要用键值，并且迭代器只能++或--，而不能+1, -1

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define x1 X1 
#define x2 X2
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair<int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) {return a < b ? a = b, 1 : 0;}
template <typename T> inline int Chkmin (T &a, T b) {return a > b ? a = b, 1 : 0;}
inline int read ()
{
	int sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = sum * 10 + ch - '0';
	return sum * fl;
}

const int Maxn = 1e6 + 100;

int N, dep[Maxn];
int e, Begin[Maxn], To[Maxn << 1], Next[Maxn << 1];
map <int, int> sum[Maxn];
pii ans[Maxn];

inline void add_edge (int x, int y)
{
	To[++e] = y;
	Next[e] = Begin[x];
	Begin[x] = e;
}

inline void dfs (int x, int fa)
{
	sum[x][dep[x]] = 1;
	ans[x] = mp(1, -dep[x]);
	for (int i = Begin[x]; i; i = Next[i])
	{
		int y = To[i];
		if (y == fa) continue;
		dep[y] = dep[x] + 1; dfs(y, x);
		if (sum[x].size() < sum[y].size())
		{
			sum[x].swap(sum[y]);
/**/		ans[x] = ans[y];
		}
		map <int, int> :: iterator it;
		for (it = sum[y].begin(); it != sum[y].end(); ++it)
		{
/**/		pii now = *it;
			sum[x][now.x] += now.y;
			ans[x] = max(ans[x], mp(sum[x][now.x], -now.x));
		}
	}
}

int main()
{
#ifdef hk_cnyali
	freopen("F.in", "r", stdin);
	freopen("F.out", "w", stdout);
#endif
	N = read();
	for (int i = 1; i < N; ++i)
	{
		int x = read(), y = read();
		add_edge (x, y);
		add_edge (y, x);
	}
	dfs(1, 0);
	for (int i = 1; i <= N; ++i)
		printf("%d\n", -ans[i].y - dep[i]);
	return 0;
}