Ciel the Commander CodeForces - 321C (树, 思维)

链接

大意: 给定n结点树, 求构造一种染色方案, 使得每个点颜色在[A,Z], 且端点同色的链中至少存在一点颜色大于端点 (A为最大颜色)

直接点分治即可, 因为最坏可以涂$2^{26}-1$个节点, 所以方案一定存在

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define pb push_back
using namespace std;
typedef long long ll;

const int N = 4e5+10, INF = 0x3f3f3f3f;
int n, sum, rt;
int mx[N], sz[N], vis[N];
vector<int> g[N];
char buf[N];

void getrt(int x, int fa) {
	mx[x]=0, sz[x]=1;
	for (int y:g[x]) if (!vis[y]&&y!=fa) {
		getrt(y,x),sz[x]+=sz[y];
		mx[x]=max(mx[x],sz[y]);
	}
	mx[x]=max(mx[x],sum-sz[x]);
	if (mx[x]<mx[rt]) rt=x;
}

void solve(int x, char c) {
	vis[x]=1, buf[x]=c;
	for (int y:g[x]) if (!vis[y]) {
		mx[rt=0]=n,sum=sz[y];
		getrt(y,0),solve(rt,c+1);
	}
}

int main() {
	scanf("%d", &n);
	REP(i,2,n) {
		int u, v;
		scanf("%d%d", &u, &v);
		g[u].pb(v), g[v].pb(u);
	}
	sum=mx[0]=n,getrt(1,0),solve(rt,'A');
	REP(i,1,n) printf("%c ", buf[i]);
	puts("");
}