Wannafly挑战赛14 - E 并查集维护线性基区间

Posted 2020-11-02 The Azure Arbitrator

给一个1-base数组{a}，有N次操作，每次操作会使一个位置无效。一个区间的权值定义为这个区间里选出一些数的异或和的最大值。求在每次操作前，所有不包含无效位置的区间的权值的最大值。

线性基删除不知道怎么维护,不妨逆向添加
然后区间连通性的维护自然要应用到并查集,每次操作mark一下当前位置,如果在操作时左边的区间已经mark过就搞它,右边同理
注意find时谁的基被插入

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<bitset>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar(‘\n‘)
#define blank putchar(‘ ‘)
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int MAXN = 1e5+11;
const double EPS = 1e-7;
typedef long long ll;
typedef unsigned long long ull;
const ll MOD = 10086; 
unsigned int SEED = 17;
const ll INF = 1ll<<60;
ll read(){
    ll x=0,f=1;register char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
struct LB{
    ll b[66];
    void clear(){
        rep(i,0,62) b[i]=0;
    }
    void insert(ll x){
        rrep(i,62,0) if(x>>i&1){
            if(b[i]) x^=b[i];
            else{
                b[i]=x;
                rrep(j,i-1,0) if(b[j]&&(b[i]>>j&1)) b[i]^=b[j];
                rep(j,i+1,62) if(b[j]>>i&1) b[j]^=b[i];
                break;
            }
        }
    }
    ll rk1(){
        ll res=0;
        rrep(i,62,0) res^=b[i];
        return res;
    }
    void merge(LB rhs){
        rep(i,0,62) if(rhs.b[i]) insert(rhs.b[i]);
    }
}B[MAXN];
struct UF{
    int p[MAXN];
    void init(int n){
        rep(i,1,n) p[i]=i;
    }
    int find(int x){
        if(x==p[x]) return x;
        // int o=p[x];
        int t=find(p[x]);
        B[t].merge(B[x]);
        return p[x]=t;
    }
    ll link(int a,int b){
        a=find(a);b=find(b);
        if(a==b) return B[a].rk1();
        p[a]=b;B[b].merge(B[a]);
        return B[b].rk1();
    }
}uf;
ll a[MAXN],n,x[MAXN],ans[MAXN];
bool vis[MAXN];
int main(){
    while(cin>>n){
        uf.init(n);
        memset(vis,0,sizeof vis);
        rep(i,1,n) a[i]=read();
        rep(i,1,n) x[i]=read();
        rep(i,1,n) B[i].clear();
        rep(i,1,n) B[i].insert(a[i]);
        ll mx=0;
        rrep(i,n,1){
            if(i==n) mx=ans[i]=a[x[n]],vis[x[n]]=1;
            else{
                ans[i]=-1;
                vis[x[i]]=1;
                if(vis[x[i]-1]&&x[i]-1>0){
                    int a=uf.find(x[i]-1);
                    mx=ans[i]=max(mx,uf.link(x[i],a));
                }
                if(vis[x[i]+1]&&x[i]+1<=n){
                    int a=uf.find(x[i]+1);
                    mx=ans[i]=max(mx,uf.link(x[i],a));
                }
                if(ans[i]==-1){
                    mx=ans[i]=max(mx,a[x[i]]);
                }
            }
            //rep(j,1,n) cout<<j<<" "<<uf.p[j]<<endl;
        }
        rep(i,1,n) println(ans[i]);
    }
    return 0;
}