【NOIP2017提高A组模拟9.7】JZOJ 计数题

【NOIP2017提高A组模拟9.7】JZOJ 计数题

题目

Description

Input

Output

Sample Input

5

2 2 3 4 5

Sample Output

8

6

Data Constraint

题解

题意

给出\(a[i]\)，有一完全图，\(i\)与\(j\)之间的边的值为\(a[i] \oplus a[j]\)（\(\oplus\)为异或的意思）

求最小生成树及方案数

题解

科普一个东西，\(n\)个点的完全图的生成树个数是\(n^{n-2}\)

这个东西叫做凯莱定理，大家可以自行了解一下

100\(\%\)

看到异或，而且要最小，且\(a[i]\)二进制做多只有30位

想到可以按照最高位往下分治，分成当前这位是0和1的两堆，然后为了取值最小，那么这两堆只能连一条

那么就找到这两堆里面异或值最小的，这是\(trie\)应用的经典问题

然后分治一位一位往下

最后把所有最小值加一起，方案数乘起来即可

Code

#include<cmath>
#include<cstdio>
#include<algorithm>
#define mod 1000000007
using namespace std;
long long n,mx,num,ans,ans1,tot,a[1000001],er[31],c1[1000001],c2[1000001];
struct node
{
	long long left,right,size;
}trie[400005];
long long read()
{
	long long res=0;char ch=getchar();
	while (ch<'0'||ch>'9') ch=getchar();
	while (ch>='0'&&ch<='9') res=(res<<1)+(res<<3)+(ch-'0'),ch=getchar();
	return res;
}
void insert(long long x)
{
	long long now=1;
	++trie[now].size;
	for (long long i=mx;i>=0;--i)
	{
		if (x&er[i])
		{
			if (trie[now].left==0) trie[now].left=++num,trie[num].left=trie[num].right=trie[num].size=0;
			now=trie[now].left;
			++trie[now].size;
		}
		else
		{
			if (trie[now].right==0) trie[now].right=++num,trie[num].left=trie[num].right=trie[num].size=0;
			now=trie[now].right;
			++trie[now].size;
		}
	}
}
long long calc(long long x)
{
	long long now=1,s=0;
	for (long long i=mx;i>=0;--i)
	{
		if (x&er[i])
		{
			if (trie[trie[now].left].size>0) now=trie[now].left;
			else s+=er[i],now=trie[now].right;
		}
		else
		{
			if (trie[trie[now].right].size>0) now=trie[now].right;
			else s+=er[i],now=trie[now].left;
		}
	}
	tot=trie[now].size;
	return s;
}
long long ksm(long long x,long long y)
{
	long long res=1;
	while (y)
	{
		if (y&1) res=res*x%mod;
		x=x*x%mod;
		y>>=1;
	}
	return res;
}
long long dg(long long l,long long r,long long d)
{
	if (r<=l) return 1;
	if (d<0) return ksm(r-l+1,r-l-1);
	long long t1=0,t2=0;
	for (long long i=l;i<=r;++i)
	{
		if (a[i]&er[d]) c1[++t1]=a[i];
		else c2[++t2]=a[i];
	}
	for (long long i=1;i<=t1;++i)
		a[l+i-1]=c1[i];
	for (long long i=1;i<=t2;++i)
		a[l+t1+i-1]=c2[i];
	long long s1=dg(l,l+t1-1,d-1),s2=dg(l+t1,r,d-1);
	long long s3=(s1*s2)%mod,s4=2147483647,s5=0;
	if (t1==0||t2==0) return s3;
	num=1;
	trie[1].left=trie[1].right=trie[1].size=0;
	for (long long i=1;i<=t1;++i)
		insert(a[l+i-1]);
	for (long long i=1;i<=t2;++i)
	{
		long long sum=calc(a[l+t1+i-1]);
		if (sum<s4)  s4=sum,s5=tot;
		else if (sum==s4) s5=(s5+tot)%mod;
	}
	ans+=s4;
	return (s3*s5)%mod;
}
int main()
{
	freopen("jst.in","r",stdin);
	freopen("jst.out","w",stdout);
	n=read();
	for (long long i=1;i<=n;++i)
		a[i]=read(),mx=max(mx,a[i]);
	mx=log2(mx);
	er[0]=1;
	for (long long i=1;i<=31;++i)
		er[i]=er[i-1]*2%mod;
	num=1;
	ans1=dg(1,n,mx);
	printf("%lld\n%lld\n",ans,ans1);
	fclose(stdin);
	fclose(stdout);
	return 0;
}