hnoi2014米特运输

思维题  略水

我们一定要用以偏概全的眼光做这道题。。。

一根节点说明一切

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MOD 1000000009
#define D 500010
#define LL long long
#define lp(i,j,k) for(int i = j;i <= k;++i)
int n,a[D],u,v,t,cnt,head[D],mx,no;
LL res[D];
bool vis[D];
struct edge {
	int v,next;
}e[D * 2];
void adde (int u,int v) {
	e[++cnt].v = v;
	e[cnt].next = head[u];
	head[u] = cnt;
	e[++cnt].v = u;
	e[cnt].next = head[v];
	head[v] = cnt;
}
void work (int X,LL ti) {
	res[X] = ti * a[X];
	res[X] %= MOD;
	LL T = 0;
	for(int i = head[X];i != -1;i = e[i].next)
		if(!vis[e[i].v])
			++T;
	T *= ti;
	T %= MOD;
	for(int i = head[X];i != -1;i = e[i].next)
		if(!vis[e[i].v]) {
			vis[e[i].v] = 1;
			work(e[i].v,T);
		}
}
int main () {
	memset(head,-1,sizeof head);
	scanf("%d",&n);
	lp(i,1,n)
		scanf("%d",&a[i]);
	lp(i,1,n - 1) {
		scanf("%d%d",&u,&v);
		adde(u,v);
	}
	vis[1] = res[1] = a[1];
	for(int i = head[1];i != -1;i = e[i].next)
		++t;
	for(int i = head[1];i != -1;i = e[i].next) {
		vis[e[i].v] = 1;
		work(e[i].v,t);
	}
	std::sort(res + 1,res + 1 + n);
	lp(i,1,n) {
		if(res[i] != res[i - 1]) {
			if(no > mx)
				mx = no;
			no = 1;
		}
		else ++no;
	}
	if(no > mx)
		mx = no;
	printf("%d\n",n - mx);
}