线性基

含log(maxval)个数a0,a1,a2...,an,ai最高的为1位为i

其中能xor出的数与原数相同

构造

  void merg(XXJ &a,XXJ &b,XXJ &c){
      for (int i=60;i>=0;i--) a.a[i]=b.a[i];
      for (int i=60;i>=0;i--)
        if (c.a[i]){
            LL tp=c.a[i];
            for (int j=60;j>=0;j--){
            if (!(tp&(1LL<<j))) continue;
          if (!a.a[j]) {a.a[j]=tp;break;}else
            tp^=a.a[j];    
        }
      }
  }

求最大亦或和时合并线性基后由高到低枚举

      for (int j=60;j>=0;j--) if ((fin^ans.a[j])>fin) 
        fin^=ans.a[j];

寻找某数是否存在

  int find(LL num){
      for (int i=62;i>=0;i--)
        if (num&(1LL<<i))
          num^=xxj[i];
      return(num==0);    
  }

 

————————————————————

BZOJ4568

#include <cstdio>
#define LL long long 

  int next[40001],des[40001],cnt,nd[20001],dep[20001],fa[20001][21],n;

  struct XXJ{
      LL a[61];
  }xxj[20001][21],ans;
  
  void merg(XXJ &a,XXJ &b,XXJ &c){
      for (int i=60;i>=0;i--) a.a[i]=b.a[i];
      for (int i=60;i>=0;i--)
        if (c.a[i]){
            LL tp=c.a[i];
            for (int j=60;j>=0;j--){
            if (!(tp&(1LL<<j))) continue;
          if (!a.a[j]) {a.a[j]=tp;break;}else
            tp^=a.a[j];    
        }
      }
  }

  void addedge(int x,int y){
      next[++cnt]=nd[x];des[cnt]=y;nd[x]=cnt;
  }

  void dfs(int po){
      for (int p=nd[po];p!=-1;p=next[p])
        if (!dep[des[p]]){
          dep[des[p]]=dep[po]+1;
        fa[des[p]][0]=po;
        dfs(des[p]);    
      }
  }
  
  void LCA_ini(){
      for (int i=1;i<=20;i++)
        for (int j=1;j<=n;j++){
          fa[j][i]=fa[fa[j][i-1]][i-1];
        if (fa[j][i]!=0)
          merg(xxj[j][i],xxj[j][i-1],xxj[fa[j][i-1]][i-1]);        
      }
  }
  
  void getlca(int x,int y){
      if (dep[x]<dep[y]){
        int t=x;x=y;y=t;    
    }
    
    for (int i=20;i>=0;i--)
      if (dep[fa[x][i]]>=dep[y]){
          merg(ans,ans,xxj[x][i]);
          x=fa[x][i];
      }
      
    for (int i=20;i>=0;i--)
      if (fa[x][i]!=fa[y][i]){
          merg(ans,ans,xxj[x][i]);
          merg(ans,ans,xxj[y][i]);
          x=fa[x][i];y=fa[y][i];
      }  
    
    if (x!=y){
      merg(ans,ans,xxj[x][0]);
      merg(ans,ans,xxj[y][0]);
      x=fa[x][0];y=fa[y][0];
    }  
    merg(ans,ans,xxj[x][0]);
  }

  int main(){      
      int q;
      scanf("%d%d",&n,&q);
      for (int i=1;i<=n;i++) nd[i]=-1;
      for (int i=1;i<=n;i++){
        LL t;
        scanf("%lld",&t);
        if (t)
      for (int j=60;j>=0;j--)
        if (t&(1LL<<j)){
          xxj[i][0].a[j]=t;
          break;    
        }
    }
      for (int i=1;i<n;i++){
        int t1,t2;
        scanf("%d%d",&t1,&t2);
      addedge(t1,t2);addedge(t2,t1);    
    }
    
    dep[1]=1;
    dfs(1);
    LCA_ini();
    
    for (int i=1;i<=q;i++){
      int t1,t2;
      scanf("%d%d",&t1,&t2);
      for (int i=0;i<=60;i++) ans.a[i]=0;
      getlca(t1,t2);
      LL fin=0;
      for (int j=60;j>=0;j--) if ((fin^ans.a[j])>fin) 
        fin^=ans.a[j];
      printf("%lld\n",fin);    
    }
  }

 ————————————————————————

用与求非线性相关组（BZOJ4004）

#include <cstdio>
#include <algorithm>
#define LL long long 
using namespace std;

  const LL mo=1e9+7;
  int bas[501];
  struct data{
      LL a[501];
      int c;
  }a[501];

  int mycomp(const data &a,const data&b){
      return(a.c<b.c);
  }
  
  LL qpow(LL bas,int powe){
      LL ret=1;
    for (;powe;bas*=bas,bas%=mo){
      if (powe&1) ret*=bas,ret%=mo;
      powe=powe>>1;
    }
    return(ret);
  }

  int main(){      
      int n,m;
      scanf("%d%d",&n,&m);
      for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
          scanf("%lld",&a[i].a[j]);
      for (int i=1;i<=n;i++)
        scanf("%d",&a[i].c);
      sort(a+1,a+n+1,mycomp);
      
      int ans=0,cnt=0;
      for (int i=1;i<=n;i++)
      for (int j=1;j<=m;j++)
        if (a[i].a[j]){
          if (!bas[j]){
              bas[j]=i;
              ans+=a[i].c;cnt++;
              break;
          }else{
              LL tim=a[i].a[j]*qpow(a[bas[j]].a[j],mo-2)%mo;
              for (int k=j;k<=m;k++){
                a[i].a[k]-=tim*a[bas[j]].a[k]%mo;a[i].a[k]%=mo;
              a[i].a[k]+=mo;a[i].a[k]%=mo;
            }
          }    
        }
     printf("%d %d\n",cnt,ans);    
  }
  
  

 -------------------------------------------------------

BZOJ4184

按时分治后dfs时间线段树求值

#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
using namespace std;

  int cnt,next[10000001],des[10000001],xxj[101][51],ans[500001],n,m;
  
  map <int,int> mpa;
  map <int,int> mpb;
  priority_queue <int> heap;

  struct treenode{
      int l,r,lc,rc,nd;
  }tr[1000001];

  struct data{
      int l,r,num;
  }sid[500001];

  void build(int l,int r){
      tr[++cnt].l=l;tr[cnt].r=r;tr[cnt].nd=-1;
      if (l==r) return;
      
      int mid=(l+r)>>1,t=cnt;
      tr[t].lc=cnt+1;
      build(l,mid);
      tr[t].rc=cnt+1;
      build(mid+1,r);
  }
  
  void edi(int po,int l,int r,int num){
      if (tr[po].l==l&&tr[po].r==r){
        next[++cnt]=tr[po].nd;des[cnt]=num;tr[po].nd=cnt;
      return;    
    }
    
    int mid=(tr[po].l+tr[po].r)>>1;
    if (l<=mid) edi(tr[po].lc,l,min(mid,r),num);
    if (r>mid)  edi(tr[po].rc,max(mid+1,l),r,num);
  }
  
  void edi(int a[],int num){
      for (int i=30;i>=0;i--)
        if (num&(1<<i)){
          if (!a[i]) {a[i]=num;return;} else
            num^=a[i];
      }
  }
  
  int getans(int a[]){
      int ret=0;
      for (int i=30;i>=0;i--)
        if ((ret^a[i])>ret)
          ret^=a[i];
      return(ret);
  }
  
  void dfs(int po,int dep){
      memcpy(xxj[dep],xxj[dep-1],sizeof(xxj[dep]));
      for (int p=tr[po].nd;p!=-1;p=next[p])
        edi(xxj[dep],des[p]);
      if (tr[po].l==tr[po].r){
        ans[tr[po].l]=getans(xxj[dep]);    
        return;
    }
    
    dfs(tr[po].lc,dep+1);
    dfs(tr[po].rc,dep+1);
  }

  int main(){
      scanf("%d",&n);
      for (int i=1;i<=n;i++){
        int t;
        scanf("%d",&t);
      if (t>0){
          if (!mpa[t]) mpb[t]=i;
          mpa[t]++;
          heap.push(t);
      }else{
          t*=-1;
          mpa[t]--;
          if (!mpa[t]) {sid[++cnt].l=mpb[t];sid[cnt].r=i-1;sid[cnt].num=t;}
      }
    }
    while (!heap.empty()){
      int num=heap.top();heap.pop();
      if (mpa[num]){
          sid[++cnt].l=mpb[num];sid[cnt].r=n;sid[cnt].num=num;
      }    
    }
    m=cnt;
    
    cnt=0;
    build(1,n);
    cnt=0;
    for (int i=1;i<=m;i++)
      edi(1,sid[i].l,sid[i].r,sid[i].num);
      
    dfs(1,1);
    
    for (int i=1;i<=n;i++) printf("%d\n",ans[i]);  
  }