P4570 [BJWC2011]元素 线性基 + 贪心

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;
}