POJ-2010 Moo University - Financial Aid---优先队列

题目链接：

https://vjudge.net/problem/POJ-2010

题目大意：

奶牛大学：奶大招生，从C头奶牛中招收N头。它们分别得分score_i，需要资助学费aid_i。希望新生所需资助不超过F，同时得分中位数最高。求此中位数。

思路：

按照分数排序，依次枚举中位数，然后用优先队列维护该头牛作为中位数的时候左边的money最小值和右边money的最小值，最后直接找出符合条件的最大的socre就可以了

#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<set>
#include<map>
#include<cmath>
using namespace std;
typedef pair<int, int> Pair;
typedef long long ll;
const int INF = 0x3f3f3f3f;
ll T, n, m, d;
const int maxn = 1e5 + 10;
struct cow
{
    ll score, money, left, right;
    bool operator <(const cow & a)const
    {
        return score < a.score;
    }
};
cow a[maxn];
int main()
{
    priority_queue<ll>q;
    ll tot, sum = 0;
    cin >> n >> m >> tot;
    for(int i = 1; i <= m; i++)scanf("%d%d", &a[i].score, &a[i].money);
    sort(a + 1, a + m + 1);
    for(int i = 1; i <= n / 2; i++)//将前n/2头牛的money加入队列
    {
        sum += a[i].money;
        q.push(a[i].money);
    }
    for(int i = n / 2 + 1; i <= m - n / 2; i++)//枚举中位数，用优先队列保持左边最小
    {
        a[i].left = sum;
        sum += a[i].money;
        q.push(a[i].money);
        sum -= q.top();
        q.pop();
    }
    while(!q.empty())q.pop();
    sum = 0;
    for(int i = m; i > m - n / 2; i--)//将后n/2头牛的money加入队列
    {
        sum += a[i].money;
        q.push(a[i].money);
    }
    for(int i = m - n / 2; i > n / 2; i--)//枚举中位数，用优先队列保持右边最小
    {
        a[i].right = sum;
        sum += a[i].money;
        q.push(a[i].money);
        sum -= q.top();
        q.pop();
    }
    ll ans = -1;
    //for(int i = 1; i <= m; i++)cout<<a[i].score<<" "<<a[i].money<<" "<<a[i].left<<" "<<a[i].right<<endl;
    for(int i = m - n / 2; i > n / 2; i--)
    {
        if(a[i].left + a[i].right + a[i].money <= tot)
        {
            ans = a[i].score;
            break;
        }
    }
    cout<<ans<<endl;
    return 0;
}