Social Infrastructure Information Systems Division, Hitachi Programming Contest 2020 C题题解

首先，我们将题目理解成若\(i\)与\(j\)距离恰好为\(3\)，则不可能\(p_i \equiv p_j \equiv 1 \space or \space 2 (\bmod 3)\)。这就相当于我们要构造一个大小为\([\frac{n + 1}{3}]\)的点集\(A_2\)，用来放所有模3余2的数，再构造一个大小为\([\frac{n + 2}{3}]\)的点集\(A_1\)，用来放所有模3余1的数。需要满足这两个集合交集为空，且若\(i\)与\(j\)距离为\(3\)，则它们不在同一个集合内。

发现距离为\(3\)的点连成的图性质是不好的，唯一好的性质就是：它是二分图。我们不妨大胆地把条件加强为：构造这样的点集\(A_1, A_2\)，使得它们全部再二分图的同一边。

我们把树按深度奇偶染色，不妨设染一种颜色的点数为\(X\)，另一种为\(Y\)，且\(X \leq Y\)。

若\(X \ge [\frac{n + 1}{3}]\)且\(Y \ge [\frac{n + 2}{3}]\)，我们就在\(X\)中任意取\([\frac{n + 1}{3}]\)个点作为\(A_2\)，在\(Y\)中任意取\([\frac{n + 2}{3}]\)个点作为\(A_1\)即可。

否则我们容易证明\(Y \ge [\frac{n + 1}{3}] + [\frac{n + 2}{3}]\)，在\(Y\)中任意找两个集合的点即可。

代码如下：

#include <bits/stdc++.h>
using namespace std;

const int N = 200005;

template <class T>
void read (T &x) {
	int sgn = 1;
	char ch;
	x = 0;
	for (ch = getchar(); (ch < '0' || ch > '9') && ch != '-'; ch = getchar()) ;
	if (ch == '-') ch = getchar(), sgn = -1;
	for (; '0' <= ch && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	x *= sgn;
}
template <class T>
void write (T x) {
	if (x < 0) putchar('-'), write(-x);
	else if (x < 10) putchar(x + '0');
	else write(x / 10), putchar(x % 10 + '0');
}

struct edge {
	int to, nxt;
} tree[N << 1];
int n, col[N], head[N], perm[N], cnt = 0, cnt0 = 0, cnt1 = 0;
void addedge (int u, int v) {
	edge e = {v, head[u]};
	tree[head[u] = cnt++] = e;
} 

void dfs (int u, int fa) {
	for (int i = head[u]; ~i; i = tree[i].nxt) {
		int v = tree[i].to;
		if (v != fa) {
			col[v] = col[u] ^ 1;
			dfs(v, u);
		}
	}
}

int main () {
	read(n);
	for (int i = 1; i <= n; i++) head[i] = -1;

	for (int i = 1; i < n; i++) {
		int u, v;
		read(u), read(v);
		addedge(u, v), addedge(v, u);
	}
	col[1] = 0, dfs(1, 0);

	for (int i = 1; i <= n; i++) {
		if (col[i] == 0) cnt0++;
		else cnt1++;
	}

	if (cnt0 > cnt1) {
		for (int i = 1; i <= n; i++) col[i] ^= 1;
		swap(cnt0, cnt1);
	}
	int bound0 = n / 3, bound1 = (n + 1) / 3, bound2 = (n + 2) / 3;
	if (cnt0 >= bound1 && cnt1 >= bound2) {
		int tot0 = 0, tot1 = 0, tot2 = 0;
		for (int i = 1; i <= n; i++) {
			if (col[i] == 0) {
				if (tot2 >= bound1) perm[i] = 3 * ++tot0;
				else perm[i] = 3 * ++tot2 - 1;
			}
			else {
				if (tot1 >= bound2) perm[i] = 3 * ++tot0;
				else perm[i] = 3 * ++tot1 - 2;
			}
		}
	}
	else {
		int tot0 = 0, tot1 = 0, tot2 = 0;
		for (int i = 1; i <= n; i++) {
			if (col[i] == 0) perm[i] = 3 * ++tot0;
			else {
				if (tot1 < bound2) perm[i] = 3 * ++tot1 - 2;
				else if (tot2 < bound1) perm[i] = 3 * ++tot2 - 1;
				else perm[i] = 3 * ++tot0;
			}
		}
	}
	for (int i = 1; i <= n; i++) write(perm[i]), putchar(' ');
	putchar('\n');
	return 0;
}