【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
%lld
",ans,ans1);
fclose(stdin);
fclose(stdout);
return 0;
}