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bzoj 3166 [Heoi2013]Alo 可持久化字典树

2016-08-07 12:56 477 查看
次大值可以用过二分加rmq方法求出,左右边求出后会形成两个区间,然后直接上可持久化字典树。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn=50005;
int root[maxn];
int cnt;
int fa[50];
namespace  Trie{
int sum[maxn*32];
int ch[maxn*32][2];
void init(){
memset(ch,0,sizeof(ch));
memset(sum,0,sizeof(sum));
}
void insert(int pre,int no,int val){
for(int i=30;i>=0;i--){
ch[no][0]=ch[pre][0];
ch[no][1]=ch[pre][1];
sum[no]=sum[pre]+1;
int w=(val>>i)&1;
ch[no][w]=++cnt;
no=ch[no][w];
pre=ch[pre][w];
}
sum[no]=sum[pre]+1;
}
int query(int pre,int no,int val){
int s=0;
for(int i=30;i>=0;i--){
int w=(val>>i)&1;
if(sum[ch[no][w^1]]-sum[ch[pre][w^1]]>0){
s+=fa[i];
w=w^1;
}
no=ch[no][w];
pre=ch[pre][w];
}
return s;
}
}
int maxl[maxn][16];
void S_table(int n)
{
int l = int(log((double)n)/log(2.0));
for (int j=1;j<=l;j++)
{
for (int i=1; i + (1 << (j-1) ) - 1 <=n;++i)
{
maxl[i][j] = max(maxl[i][j-1], maxl[i + (1 << (j-1) )][j-1]);
}
}
}

int rmq(int l, int r)
{
int k = int(log((double)(r-l+1))/log(2.0));
int a1 = max(maxl[l][k], maxl[r - (1<<k) + 1][k]);
return a1;
}
int a[maxn];
int main()
{
for(int i=0;i<=30;i++)fa[i]=1<<i;
int n;
cnt=0;
cin>>n;
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
root[i]=++cnt;
Trie::insert(root[i-1],root[i],a[i]);
maxl[i][0]=a[i];
}
S_table(n);
int ans=0;
for(int i=1;i<=n;i++){
int le=1,ri=i-1;
while(le<=ri){
int mid=(le+ri)/2;
if(rmq(mid,i-1)>a[i]) le=mid+1;
else ri=mid-1;
}
int l1=ri;
int l2=0;
if(ri>1){
le=1;
ri=l1-1;
while(le<=ri){
int mid=(le+ri)/2;
if(rmq(mid,l1-1)>a[i]) le=mid+1;
else ri=mid-1;
}
l2=ri;
}
le=i+1,ri=n;
while(le<=ri){
int mid=(le+ri)/2;
if(rmq(i+1,mid)>a[i]) ri=mid-1;
else le=mid+1;
}
int r1=le;
int r2=n+1;
if(r1<n){
le=r1+1,ri=n;
while(le<=ri){
int mid=(le+ri)/2;
if(rmq(r1+1,mid)>a[i]) ri=mid-1;
else le=mid+1;
}
r2=le;
}
if(l1>=1){
ans=max(ans,Trie::query(root[l2],root[r1-1],a[i]));
}
if(r1<=n){
ans=max(ans,Trie::query(root[l1],root[r2-1],a[i]));
}
}
cout<<ans<<endl;
}
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