[SCOI2016]幸运数字

算简单的了，直接拿线性基维护路径异或然后倍增跳合并就行。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iostream>
using namespace std;
#define re register
#define ll long long
inline int gi(){
    int f=1,sum=0;char ch=getchar();
    while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
    return f*sum;
}
inline ll gl(){
    ll f=1,sum=0;char ch=getchar();
    while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
    return f*sum;
}
const int N=20010;
struct xxj{
    ll p[62];
    void insert(ll &x){
        for(int i=61;~i;i--)
            if((x>>i)&1){
                if(!p[i]){p[i]=x;return;}
                x^=p[i];
            }
        return;
    }
    void merge(xxj &b){
        for(int i=61;~i;i--)
            if(b.p[i])insert(b.p[i]);
        return;
    }
}xj[N][17],ans;//15
int f[N][17],n,Q,front[N],cnt;
struct node{int to,nxt;}e[400010];
void Add(int u,int v){e[++cnt]=(node){v,front[u]};front[u]=cnt;}
ll g[N];
int dep[N];
void dfs(int u,int ff){
    f[u][0]=ff;xj[u][0].insert(g[u]);dep[u]=dep[ff]+1;
    for(int i=front[u];i;i=e[i].nxt){
        int v=e[i].to;if(v==ff)continue;
        dfs(v,u);
    }
}
void lca(int u,int v){
    if(dep[u]<dep[v])swap(u,v);
    memset(ans.p,0,sizeof(ans.p));
    for(int i=15;~i;i--)
        if(dep[u]-(1<<i)>=dep[v]){
            ans.merge(xj[u][i]);
            u=f[u][i];
        }
    if(u==v){
        ans.merge(xj[u][0]);
        return;
    }
    for(int i=15;~i;i--)
        if(f[u][i]!=f[v][i]){
            ans.merge(xj[u][i]);ans.merge(xj[v][i]);
            u=f[u][i],v=f[v][i];
        }
    ans.merge(xj[u][0]);ans.merge(xj[v][0]);
    ans.merge(xj[f[u][0]][0]);
}
int main(){
    n=gi();Q=gi();
    for(int i=1;i<=n;i++)g[i]=gl();
    for(int i=1;i<n;i++){int u=gi(),v=gi();Add(u,v);Add(v,u);}
    dfs(1,1);
    for(int j=1;j<=15;j++)
        for(int i=1;i<=n;i++){
            f[i][j]=f[f[i][j-1]][j-1];
            xj[i][j]=xj[i][j-1];
            xj[i][j].merge(xj[f[i][j-1]][j-1]);
        }
    while(Q--){
        int u=gi(),v=gi();
        lca(u,v);
        ll sum=0;
        for(int i=61;~i;i--)
            if(ans.p[i])sum=max(sum,sum^(ll)ans.p[i]);
        printf("%lld\n",sum);
    }
    return 0;
}

标签：xj,ch,数字,int,merge,ans,SCOI2016,幸运,include
来源： https://www.cnblogs.com/mle-world/p/11146016.html