Educational Codeforces Round 55 (Rated for Div. 2) C. Multi-Subject Competition （实现，贪心，排序）

C. Multi-Subject Competition
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
A multi-subject competition is coming! The competition has m different subjects participants can choose from. That‘s why Alex (the coach) should form a competition delegation among his students.

He has n candidates. For the i-th person he knows subject si the candidate specializes in and ri — a skill level in his specialization (this level can be negative!).

The rules of the competition require each delegation to choose some subset of subjects they will participate in. The only restriction is that the number of students from the team participating in each of the chosen subjects should be the same.

Alex decided that each candidate would participate only in the subject he specializes in. Now Alex wonders whom he has to choose to maximize the total sum of skill levels of all delegates, or just skip the competition this year if every valid non-empty delegation has negative sum.

(Of course, Alex doesn‘t have any spare money so each delegate he chooses must participate in the competition).

Input
The first line contains two integers n and m (1≤n≤105, 1≤m≤105) — the number of candidates and the number of subjects.

The next n lines contains two integers per line: si and ri (1≤si≤m, ?104≤ri≤104) — the subject of specialization and the skill level of the i-th candidate.

Output
Print the single integer — the maximum total sum of skills of delegates who form a valid delegation (according to rules above) or 0 if every valid non-empty delegation has negative sum.

Examples
inputCopy
6 3
2 6
3 6
2 5
3 5
1 9
3 1
outputCopy
22
inputCopy
5 3
2 6
3 6
2 5
3 5
1 11
outputCopy
23
inputCopy
5 2
1 -1
1 -5
2 -1
2 -1
1 -10
outputCopy
0
Note
In the first example it‘s optimal to choose candidates 1, 2, 3, 4, so two of them specialize in the 2-nd subject and other two in the 3-rd. The total sum is 6+6+5+5=22.

In the second example it‘s optimal to choose candidates 1, 2 and 5. One person in each subject and the total sum is 6+6+11=23.

In the third example it‘s impossible to obtain a non-negative sum.

题意：
有一个竞赛，有m个科目，现在有n个学生，每一个学生只对一个科目有一个水平值。现在要求选出一些学生，要求这些学生参加的科目中，每一个科目的学生数量应该相同。 现在问你选可以选择出的最大水平值sum和是多少？
思路：

把学生信息按照擅长的科目加入到一个vector中，然后对每一个科目的Vector进行排序，维护一个数组 mx[i] 代表所有科都（如果人数小于i就贡献为0，可以忽略）选择了 i 个人的水平sum和是多少。因为有负的水平，所以我们在处理的时候，如果一个项目中学生的水平sum和小于0的时候，就结束这个项目的维护，以此保证答案最优。

细节见代码：

#include<bits/stdc++.h>
using namespace std;

#define fore(i, l, r) for(int i = int(l); i < int(r); i++)
#define sz(a) int((a).size())

int n, m;
vector<int> s, r;

inline bool read() {
    if(!(cin >> n >> m))
        return false;
    s.assign(n, 0);
    r.assign(n, 0);

    fore(i, 0, n) {
        assert(cin >> s[i] >> r[i]);
        s[i]--;
    }
    return true;
}

vector< vector<int> > subs;

inline void solve() {
    subs.assign(m + 1, vector<int>());

    fore(i, 0, n)
        subs[s[i]].push_back(r[i]);

    fore(id, 0, sz(subs)) {
        sort(subs[id].begin(), subs[id].end());
        reverse(subs[id].begin(), subs[id].end());
    }

    vector<int> mx(n + 5, 0);
    fore(id, 0, sz(subs)) {
        int curSum = 0;
        fore(i, 0, sz(subs[id])) {
            curSum += subs[id][i];
            if(curSum < 0)
                break;

            mx[i + 1] += curSum;
        }
    }

    cout << *max_element(mx.begin(), mx.end()) << endl;
}

int main() {
#ifdef _DEBUG
    freopen("input.txt", "r", stdin);
    int tt = clock();
#endif
    cout << fixed << setprecision(15);

    if(read()) {
        solve();

#ifdef _DEBUG
        cerr << "TIME = " << clock() - tt << endl;
        tt = clock();
#endif
    }
    return 0;
}

原文地址：https://www.cnblogs.com/qieqiemin/p/11185074.html