牛客提高D4t3 清新题

分析

树上从下往上线性基合并即可

并不需要启发式/xyx

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
#define int long long
int n,m,belong[100100][31],ans[100100],val[100100],res[100100],siz[100100];
int id[100100];
vector<int>v[100100];
inline void ins(int wh,int x){
    int i,j,k;
    for(j=30;j>=0;j--)
        if((1ll<<j)&x){
          if(!belong[wh][j]){
              belong[wh][j]=x;
              break;
          }
          x^=belong[wh][j];
        }
    return;
}
inline void mer(int x,int y){
    int i,j,k;
    for(j=30;j>=0;j--)
      if(belong[y][j])ins(x,belong[y][j]);
    return;
}
inline void dfs(int x,int fa){
    int i,j,k;
    ins(id[x],val[x]);
    for(i=0;i<v[x].size();i++)
      if(v[x][i]!=fa){
          dfs(v[x][i],x);
          mer(x,v[x][i]);
      }
    for(i=30;i>=0;i--)
      if((ans[x]^belong[id[x]][i])>ans[x])ans[x]=(ans[x]^belong[id[x]][i]);
}
signed main(){
    int i,j,k;
    scanf("%lld",&n);
    for(i=1;i<n;i++){
      int x,y;
      scanf("%lld%lld",&x,&y);
      v[x].push_back(y);
      v[y].push_back(x);
    }
    for(i=1;i<=n;i++)scanf("%lld",&val[i]),id[i]=i;
    dfs(1,0);
    scanf("%lld",&m);
    while(m--){
      int x;
      scanf("%lld",&x);
      printf("%lld\n",ans[x]);
    }
    return 0;
}

标签：D4t3,int,belong,100100,id,牛客,ans,include,清新
来源： https://www.cnblogs.com/yzxverygood/p/11405787.html