JZOJ5913 林下风气

Posted 2020-11-30 dftmr

询问最大值与最小值之差为k好像比较困难，所以可以把转换成求最大值与最小值之差小于等于k的问题，等于k的数量=小于等于k的数量-小于等于k-1的数量

这样可以在树上进行DFS，时间复杂度O(n^2)

#include<cstdio>
using namespace std;
typedef long long ll;
const ll mod=19260817;
int head[3335],len,n,k,x,y;
ll val[3335],ans;
struct EDGE{
	int to,next;
}edge[7005];
void add(int x,int y){
	++len;
	edge[len].to=y;
	edge[len].next=head[x];
	head[x]=len;
}
ll dfs(int u,int fa,int root,int k){
	ll mul=1;
	for (register int i=head[u];i;i=edge[i].next){
		int v=edge[i].to;
		if (v!=fa&&val[root]>=val[v]&&(val[root]!=val[v]||root<v)&&val[root]-val[v]<=k){
			mul=1ll*mul*(dfs(v,u,root,k)+1);
			mul%=mod;
		} 
	}
	return mul;
}
int main(){
	scanf ("%d%d",&n,&k);
	for (register int i=1;i<=n;++i) scanf ("%d",&val[i]);
	for (register int i=1;i<n;++i){
		scanf ("%d%d",&x,&y);
		add(x,y);add(y,x);
	}
	for (register int i=1;i<=n;++i){
		if (k==0) ans=(ans+dfs(i,0,i,k))%mod;
		else ans=(ans+dfs(i,0,i,k)-dfs(i,0,i,k-1)+mod)%mod;
	}
	printf ("%lld
",ans);
	return 0;
}