Xor && 线性基练习

#include <cstdio>
#include <cstring>
const int Len=31;
const int Maxn=100100;
int cnt,Ans,b,x,n;
inline int Max(int x,int y) {return x>y?x:y;}
struct Node {int next[2];}Tree[Maxn*Len];
void Insert(int x)
{
    int Now=0; bool k;
    for (int i=Len;i>=0;i--)
    {
        k=x&(1<<i);
        if (Tree[Now].next[k]==-1) Tree[Now].next[k]=++cnt;
        Now=Tree[Now].next[k];
    }
}
int Query(int x)
{
    int Now=0,v=0; bool k;
    for (int i=Len;i>=0;i--)
    {
        k=x&(1<<i);
        if (Tree[Now].next[!k]!=-1) k=!k;
        v=v|(k<<i);
        Now=Tree[Now].next[k];
    }
    return v;
}
int main()
{
    // freopen("c.in","r",stdin);
    while (scanf("%d",&n)!=EOF)
    {
        Ans=cnt=0; memset(Tree,-1,sizeof(Tree));
        for (int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            Insert(x);
            Ans=Max(Ans,x^Query(x));
        }
        printf("%d
",Ans);
    }
    return 0;
}

在n个数取两个Xor最大。用0-1Trie.

#include <cstdio>
#include <algorithm>
#define LL long long
using namespace std;
LL n,Ans,Base[1010];
struct Node{LL a,b;}A[1010];
inline bool cmp(Node A,Node B) {return A.b>B.b;}
 
int main()
{
    scanf("%lld",&n);
    for (LL i=1;i<=n;i++) scanf("%lld%lld",&A[i].a,&A[i].b);
    sort(A+1,A+n+1,cmp); Ans=0;
    for (LL i=1;i<=n;i++)
    {
        for (LL j=63;j>=0;j--)
            if (A[i].a>>j&1)
            {
                if (!Base[j])
                {
                    Base[j]=A[i].a;
                    break;
                }
                A[i].a^=Base[j];
            }
        if (A[i].a) Ans+=A[i].b;
    }
    printf("%lld
",Ans);
    return 0;
}

若子集异或为0，则集合一定不为线性基。从大到小排序，贪心即可。

#include <cstdio>
#include <iostream>
using namespace std;
#define LL long long
const LL Maxn=100100;
LL a[Maxn],Kase,n,Ans,k,q,Bin[100];
inline void Swap(LL &x,LL &y) {LL t=x;x=y;y=t;}
inline LL Get(LL Row,LL k)
{
    if (Row<n)
    {if (k==1) return 0; else k--;}
    if (k>=Bin[Row]) return -1;
    LL Ret=0;
    for (int i=1;i<=Row;i++)
        if (k&Bin[Row-i]) Ret^=a[i];
    return Ret;
}
int main()
{
    scanf("%I64d",&Kase);
    Bin[0]=1; for (int i=1;i<=60;i++) Bin[i]=Bin[i-1]<<1;
    for (LL kase=1;kase<=Kase;kase++)
    {
        printf("Case #%I64d:
",kase);
        scanf("%I64d",&n);
        for (LL i=1;i<=n;i++) scanf("%I64d",&a[i]);
        LL Now=0;
        for (LL i=Bin[60];i;i>>=1)
        {
            LL j=Now+1;
            while (!(i&a[j])&&j<=n) j++;
            if (j==n+1) continue;
            ++Now; Swap(a[Now],a[j]);
            for (j=1;j<=n;j++)
            {
                if (j==Now) continue;
                if (a[j]&i) a[j]=a[j]^a[Now];
            }
        }        
        scanf("%I64d",&q);
        for (LL i=1;i<=q;i++)
        {
            scanf("%I64d",&k);
            printf("%I64d
",Get(Now,k));
        }
    }
    return 0;
}

求第K大的Xor和，求出线性基，把K转为二进制在Xor就可以了

#include <cstdio>
#include <algorithm>
#define LL long long
using namespace std;
const LL Mod=1000000009;
const LL Maxn=1010;
struct Node{LL b[Maxn],c;}a[Maxn];
LL Base[Maxn],cnt,Ans,n,m;
inline LL Pow(LL x,LL y)
{
    LL Ret=1;
    while (true)
    {
        if (y&1) Ret=(Ret*x)%Mod;
        x=(x*x)%Mod; y>>=1; 
        if (y==0) break;
    }
    return Ret;
}
inline bool cmp(Node A,Node B) {return A.c<B.c;}
int main()
{
    scanf("%lld%lld",&n,&m);
    for (LL i=1;i<=n;i++)
        for (LL j=1;j<=m;j++) scanf("%lld",&a[i].b[j]);
    for (LL i=1;i<=n;i++) scanf("%lld",&a[i].c);
    sort(a+1,a+n+1,cmp);
    for (LL i=1;i<=n;i++)
    {
        bool flag=false;
        for (LL j=1;j<=m;j++)
            if (a[i].b[j])
            {
                if (!Base[j])
                {
                    Base[j]=i; flag=true;
                    break;
                }
                LL t=Mod-(a[i].b[j]*Pow(a[Base[j]].b[j],Mod-2))%Mod;
                for (LL k=j;k<=m;k++)
                    a[i].b[k]=(a[i].b[k]+(t*a[Base[j]].b[k])%Mod)%Mod;
            }
        if (flag) Ans+=a[i].c,cnt++;
    }
    printf("%lld %lld
",cnt,Ans);
    return 0;
}

即求出线性基，然后贪心取即可。

#include <cstdio>
#include <cstring>
#define LL long long
inline void Get_Int(LL & x)
{
    char ch=getchar(); x=0;
    while (ch<'0' || ch>'9') ch=getchar();
    while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();}
}
inline void Swap(LL &x,LL &y) {LL t=x;x=y;y=t;}
inline LL Max(LL x,LL y) {return x>y?x:y;}
inline LL Min(LL x,LL y) {return x>y?y:x;}
//==================================================
const LL Maxn=20010;
LL head[Maxn],father[Maxn][20],Dep[Maxn],n,m,Bin[70],x,u,v,cnt,Sum;
bool vis[Maxn];
struct Edge{LL to,next;}edge[Maxn<<2];
struct Base
{
    LL a[70];
    inline void Clr() {memset(a,0,sizeof(a));}
    inline void Insert(LL x) 
    {
        for (LL i=60;i>=0;i--)
            if (Bin[i]&x) 
            {
                if (!a[i]) {a[i]=x; return;}
                x^=a[i];
            }
    }
}; 
Base f[Maxn][20],Ans;
inline void Add(LL u,LL v) {edge[cnt].to=v;edge[cnt].next=head[u];head[u]=cnt++;}
void Dfs(LL u)
{
    vis[u]=true;
    for (LL i=head[u];i!=-1;i=edge[i].next) 
        if (!vis[edge[i].to])
        {
            father[edge[i].to][0]=u;
            Dep[edge[i].to]=Dep[u]+1;
            Dfs(edge[i].to);
        }
}
inline void Init()
{
    for (LL j=1;j<=15;j++)
        for (LL i=1;i<=n;i++)
        {
            father[i][j]=father[father[i][j-1]][j-1];
            for (LL k=60;k>=0;k--) if (f[i][j-1].a[k])f[i][j].Insert(f[i][j-1].a[k]);
            for (LL k=60;k>=0;k--) if (f[father[i][j-1]][j-1].a[k])f[i][j].Insert(f[father[i][j-1]][j-1].a[k]);
        }
}
void Solve(LL u,LL v)
{
    if (Dep[u]<Dep[v]) Swap(u,v);
    for (LL i=15;i>=0;i--)
        if (Dep[father[u][i]]>=Dep[v])
        {
            for (LL k=60;k>=0;k--) if (f[u][i].a[k]) Ans.Insert(f[u][i].a[k]);
            u=father[u][i];
        }
    if (u==v) 
    {
        for (LL k=60;k>=0;k--) if (f[u][0].a[k]) Ans.Insert(f[u][0].a[k]);
        return;
    }
     
    for (LL i=15;i>=0;i--)
        if (father[u][i]!=father[v][i])
        {
            for (LL k=60;k>=0;k--) if (f[u][i].a[k]) Ans.Insert(f[u][i].a[k]);
            for (LL k=60;k>=0;k--) if (f[v][i].a[k]) Ans.Insert(f[v][i].a[k]);
            u=father[u][i],v=father[v][i];
        }
    for (LL k=60;k>=0;k--) if (f[u][0].a[k])Ans.Insert(f[u][0].a[k]);
    for (LL k=60;k>=0;k--) if (f[v][0].a[k])Ans.Insert(f[v][0].a[k]);
    for (LL k=60;k>=0;k--) if (f[father[u][0]][0].a[k])Ans.Insert(f[father[u][0]][0].a[k]);
}
 
int main()
{
    Bin[0]=1; for (LL i=1;i<=60;i++) Bin[i]=Bin[i-1]<<1;
    Get_Int(n),Get_Int(m);
    for (LL i=1;i<=n;i++)
    {
        Get_Int(x);
        f[i][0].Insert(x);
    }
    memset(head,-1,sizeof(head));
    for (LL i=1;i<n;i++)
    {
        Get_Int(u),Get_Int(v);
        Add(u,v),Add(v,u);
    }
    father[1][0]=1; Dep[1]=1;
    memset(vis,false,sizeof(vis)); Dfs(1);
    Init();
    for (LL i=1;i<=m;i++)
    {
        Get_Int(u),Get_Int(v);
        Ans.Clr();
        Solve(u,v); Sum=0;
        for (LL j=60;j>=0;j--) Sum=Max(Sum,Sum^Ans.a[j]);
        printf("%lld
",Sum);
    }
    return 0;
}

倍增合并线性基，贪心求最大值即可