poj2010(Moo University - Financial Aid)优先队列

标签：acm   优先队列   

3 5 70
30 25
50 21
20 20
5 18
35 30

35

题意：给你 c头牛和每头牛的智商、花费，要你选择 n头牛使这 n头牛智商的中位数最大，前提是花费总和不能超过 f

题解：先对每头牛的智商从小到大排序，然后枚举智商的中位数，对于某个中位数i，左边和右边各取花费最小的n/2头牛。怎么取呢？先用优先队列预处理每头牛左右两边n/2头牛花费最小值即可。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;

const int maxn = 100010;

priority_queue<int> p, q;
int n, c, f;
pair<int, int> cow[maxn];
int l[maxn], r[maxn];

int main()
{
    cin >> n >> c >> f;
    for (int i = 0; i < c; ++i)
        scanf("%d%d", &cow[i].first, &cow[i].second);
    sort(cow, cow+c);

    //预处理以i为中位数时左边n/2头牛花费的最小值
    int s = 0;
    for (int i = 0; i < c; ++i)
    {
        if (p.size() == n / 2)
            l[i] = s;
        p.push(cow[i].second);
        s += cow[i].second;
        if (p.size() > n / 2)
        {
            s -= p.top();
            p.pop();
        }
    }

    //预处理以i为中位数时右边n/2头牛花费的最小值
    s = 0;
    for (int i = c-1; i >= 0; --i)
    {
        if (q.size() == n / 2)
            r[i] = s;
        q.push(cow[i].second);
        s += cow[i].second;
        if (q.size() > n / 2)
        {
            s -= q.top();
            q.pop();
        }
    }

    //从大到小枚举中位数，满足题意就退出
    for (int i = c-(n+1)/2; i >= n/2; --i)
    {
        int sum = cow[i].second + l[i] + r[i];
        if (sum <= f)
        {
            cout << cow[i].first << endl;
            return 0;
        }
    }
    cout << "-1" << endl;
    return 0;
}

poj2010(Moo University - Financial Aid)优先队列

标签：acm   优先队列   

原文地址：http://blog.csdn.net/god_weiyang/article/details/47752259