CF486D Valid Sets

枚举最大点i，统计以i为最大点有多少个连通块，由于一个连通块可能有许多个最大点，我们强制选编号最小的那个即可。

#include<bits/stdc++.h>
#define file(s) freopen(s".in","r",stdin);freopen(s".out","w",stdout);
#define P 1000000007
#define mid (l+r>>1)
#define N 1100000
#define lb(x) (x&(-x))
#define inf 999999999
#define M 1658561
#define int long long
#define mem(x) memset(x,0,sizeof(x));
using namespace std;
int n,d,a[N],dp[N],ans,rt,to[N],nxt[N],head[N],cnt;
void add(int x,int y){
	to[++cnt]=y;
	nxt[cnt]=head[x];
	head[x]=cnt;
}
void dfs(int x,int fa){
	dp[x]=1;
	for(int i=head[x];i;i=nxt[i]){
		if(to[i]==fa) continue;
		if(a[to[i]]<a[rt]-d||a[to[i]]>a[rt]||(a[to[i]]==a[rt]&&to[i]<rt)) continue;
		dfs(to[i],x);(dp[x]+=dp[x]*dp[to[i]])%=P;
	}
}
signed main(){
	int x,y;
	scanf("%lld%lld",&d,&n);
	for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
	for(int i=1;i<n;i++) scanf("%lld%lld",&x,&y),add(x,y),add(y,x);
	for(int i=1;i<=n;i++) rt=i,dfs(i,0),(ans+=dp[i])%=P;
	printf("%lld
",ans);
	return 0;
}