标签:name 思想 down 端点 contest 修改 size ini ios
https://codeforces.com/contest/587/problem/E
一个序列,
1区间异或操作
2查询区间子集异或种类数
题解
解题思路大同小异,都是利用异或的性质进行转化,std和很多网友用的都是差分的思想,用两棵线段树
第一棵维护差分序列上的线性基,第二棵维护原序列的异或区间和,两者同时进行修改
考虑两个序列 $(a,b)(d,e)$,按照std的想法,应该是维护$(0 \^ a,a \^ b)(0 \^ d,d \^ e)$ 然后合并首尾变成$(0 \^ a,a \^ b,b \^ d,d \^ e)$
但由于异或的性质,我们直接每个区间保存下区间左端点原来的信息,
直接先插入两个序列的线性基,然后新头部的异或和即可,也就是$(0 \^a,a \^b,a \^ d,d \^e)$
写起来更加轻松
#include <bits/stdc++.h> #define endl ‘\n‘ #define ll long long #define ull unsigned long long #define fi first #define se second #define mp make_pair #define pii pair<ll,ll> #define all(x) x.begin(),x.end() #define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0) #define rep(ii,a,b) for(register int ii=a;ii<=b;++ii) #define per(ii,a,b) for(register int ii=b;ii>=a;--ii) #define forn(ii,x) for(int ii=head[x];ii;ii=e[ii].next) #define show(x) cout<<#x<<"="<<x<<endl #define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl #define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl #define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl #define show5(v,w,x,y,z) cout<<#v<<" "<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl #define showa(a,b) cout<<#a<<‘[‘<<b<<"]="<<a[b]<<endl using namespace std; const int maxn=2e5+10,maxm=2e5+10; const int INF=0x3f3f3f3f; const ll mod=1e9+7; const double PI=acos(-1.0); //heads int casn,n,m,k; class segtree{public: #define nd node[now] #define ndl node[now<<1] #define ndr node[now<<1|1] struct segnode{ int l,r,flag,val; int d[32]; inline void init(){val=flag=0;memset(d,0,sizeof d);} inline void insert(ll x){ for(register int i=30;x&&i>=0;--i) if(x&(1ll<<i)){ if(!d[i]) {d[i]=x;return;} else x^=d[i]; } } int count(){int ans=0;per(i,0,30) if(d[i])ans++; return ans;} void update(int x){val^=x;flag^=x;} }node[maxn<<2|3]; inline segnode marge(segnode &a,segnode b)const { segnode ans;ans.init(); per(i,0,30) ans.insert(a.d[i]),ans.insert(b.d[i]); ans.insert(a.val^b.val); ans.val=a.val; ans.l=a.l,ans.r=b.r; return ans; } inline void down(int now){ if(nd.flag){ ndl.update(nd.flag);ndr.update(nd.flag); nd.flag=0; } } void maketree(int s,int t,int now=1){ nd.l=s,nd.r=t;nd.init(); if(s==t) {cin>>nd.val;return ;} maketree(s,(s+t)/2,now<<1); maketree((s+t)/2+1,t,now<<1|1); nd=marge(ndl,ndr); } void update(int s,int t,int x,int now=1){ if(s<=nd.l&&t>=nd.r) {nd.update(x);return;} down(now); if(s<=ndl.r) update(s,t,x,now<<1); if(t>ndl.r) update(s,t,x,now<<1|1); nd=marge(ndl,ndr); } segnode query(int s,int t,int now=1){ if(s<=nd.l&&t>=nd.r) { if(s==nd.l) { segnode x;x.init(); return marge(x,nd); }else return nd; } down(now); segnode ans;ans.init(); if(s<=ndl.r) ans=marge(ans,query(s,t,now<<1)); if(t>ndl.r) ans=marge(ans,query(s,t,now<<1|1)); nd=marge(ndl,ndr); return ans; } }tree; int main() { IO; cin>>n>>m; register int a,b,c,d; tree.maketree(1,n); while(m--){ cin>>a>>b>>c; if(a==1){ cin>>d;tree.update(b,c,d); }else cout<<(1<<tree.query(b,c).count())<<endl; } }
CodeForces 587 E.Duff as a Queen 线段树动态维护区间线性基
标签:name 思想 down 端点 contest 修改 size ini ios
原文地址:https://www.cnblogs.com/nervendnig/p/10739751.html