题意

给定n个物品，每个物品有一个编号和价值，问如何取使得拿到的物品价值总和最大，并且取得物品的编号的子集异或和不能为0。

思路

这是个贪心，我们先按照价值从大到小排序，然后贪心地取，如果当前要取的物品的编号和之前取的存在异或为0的情况，我们就丢弃这个物品，否则加入。判断异或为0可以用线性基来做。
具体证明参考

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>


using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost ios::sync_with_stdio(false);cin.tie(0)
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);


const ll oo = 1ll<<17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}

inline void cmax(int &x,int y){if(x<y)x=y;}
inline void cmax(ll &x,ll y){if(x<y)x=y;}
inline void cmin(int &x,int y){if(x>y)x=y;}
inline void cmin(ll &x,ll y){if(x>y)x=y;}


/*-----------------------showtime----------------------*/
            const int maxn = 1009;
            struct node{
                ll id;
                int val;
            }a[maxn];
            bool cmp(node a,node b){
                return a.val > b.val;
            }
            ll p[109];
            bool check(ll x){
                for(int i=60; i>=0; i--) {
                    if((x & (1ll << i) ) > 0)
                    {
                        if(p[i] >-1) x ^= p[i];
                        else {
                            p[i] = x;
                            return true;
                        }
                    }
                }
                return false;
            }

int main(){
            memset(p, -1, sizeof(p));
            int n;  scanf("%d", &n);
            for(int i=1; i<=n; i++) {
                scanf("%lld%d", &a[i].id, &a[i].val);
            }
            sort(a+1, a+1+n,cmp);
            int sum = 0;
            for(int i=1; i<=n; i++) {
                if(check(a[i].id)) sum += a[i].val;
            }
            printf("%d\n", sum);

            return 0;
}

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