优先队列 POJ 2010

题意：

有个学校，从c个学生中挑n个学生，要求每个学生的资助金总和<=f

求符合条件的选择方案中中学生成绩中位数的最大值

题解:

有一个lower数组，记录当第i个人是中位数时候，前面的人所需要的资金的最小值，同理有一个upper数组

lower[i] = q.size() == half ? total : 0x3f3f3f3f;

total是当前最小值

if (q.size() > half)
{
total -= q.top(); q.pop();
}

因为这句话保证了当前最小

#include <iostream>
#include <algorithm>
#include <queue>
#include<stdio.h>
#include<string.h>
using namespace std;
#define MAX_COW 100000 + 16
int N, C;
long long F;
pair<int, int> cow[MAX_COW];
int lower[MAX_COW], upper[MAX_COW];
int main()
{
cin >> N >> C >> F;
int half = N / 2;
for (int i = 0; i < C; ++i)
{
scanf("%d%d",&cow[i].first,&cow[i].second);
}

sort(cow, cow + C);
{
int total = 0;
priority_queue<int> q;
for (int i = 0; i < C; ++i)
{
lower[i] = q.size() == half ? total : 0x3f3f3f3f;
q.push(cow[i].second);
total += cow[i].second;
if (q.size() > half)
{
total -= q.top(); q.pop();
}}
}

{
int total = 0;
priority_queue<int> q;
for (int i = C - 1; i >= 0; --i)
{
upper[i] = q.size() == half ? total : 0x3f3f3f3f;
q.push(cow[i].second);
total += cow[i].second;
if (q.size() > half)
{
total -= q.top(); q.pop();
}}
}

int result = -1;
for (int i = C - 1; i >= 0; --i)
{
if (lower[i] + cow[i].second + upper[i] <= F)
{
result = cow[i].first;
break;
}
}

printf("%d\n",result);
return 0;
}