链接:

https://codeforces.com/contest/1230/problem/D

题意:

Marcin is a coach in his university. There are n students who want to attend a training camp. Marcin is a smart coach, so he wants to send only the students that can work calmly with each other.

Let‘s focus on the students. They are indexed with integers from 1 to n. Each of them can be described with two integers ai and bi; bi is equal to the skill level of the i-th student (the higher, the better). Also, there are 60 known algorithms, which are numbered with integers from 0 to 59. If the i-th student knows the j-th algorithm, then the j-th bit (2j) is set in the binary representation of ai. Otherwise, this bit is not set.

Student x thinks that he is better than student y if and only if x knows some algorithm which y doesn‘t know. Note that two students can think that they are better than each other. A group of students can work together calmly if no student in this group thinks that he is better than everyone else in this group.

Marcin wants to send a group of at least two students which will work together calmly and will have the maximum possible sum of the skill levels. What is this sum?

思路:

由题意可得, 组合内必须有两个相同的, 考虑所有拥有两个或两个以上相同a的集合.
这些集合可以组成一个大集合.同时其他值只要不存在有这些集合共有的值即可.
判定过程使用位运算可以优化到O(n)(没试过)

代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 7e3+10;

struct Node
{
    LL a, b;
}node[MAXN];
map<LL, pair<int, LL> > Mp;
LL Id[MAXN], Val[MAXN];
int n, cnt = 0;

bool Check(LL a, LL b)
{
    while (b)
    {
        if (((a&1) == 0) && ((b&1) == 1))
            return false;
        a >>= 1;
        b >>= 1;
    }
    return true;
}

int main()
{
    cin >> n;
    for (int i = 1;i <= n;i++)
        cin >> node[i].a;
    for (int i = 1;i <= n;i++)
        cin >> node[i].b;
    for (int i = 1;i <= n;i++)
    {
        Mp[node[i].a].first++;
        Mp[node[i].a].second += node[i].b;
    }
    LL maxa = 0, maxb = 0;
    for (auto x:Mp)
    {
        if (x.second.first > 1)
        {
            Id[++cnt] = x.first;
            Val[cnt] = x.second.second;
        }
    }
    if (cnt == 0)
    {
        puts("0");
        return 0;
    }
    LL res = 0;
    for (int i = 1;i <= n;i++)
    {
        for (int j = 1;j <= cnt;j++)
        {
            if (node[i].a == Id[j])
                break;
            if (Check(Id[j], node[i].a))
            {
                res += node[i].b;
                break;
            }
        }
    }
    for (int i = 1;i <= cnt;i++)
        res += Val[i];
    cout << res << endl;

    return 0;
}

原文地址：https://www.cnblogs.com/YDDDD/p/11621900.html