poj 2010 Moo University - Financial Aid
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 7961 | Accepted: 2321 |
Description
Not wishing to admit dumber-than-average cows, the founders created an incredibly precise admission exam called the Cow Scholastic Aptitude Test (CSAT) that yields scores in the range 1..2,000,000,000.
Moo U is very expensive to attend; not all calves can afford it.In fact, most calves need some sort of financial aid (0 <= aid <=100,000). The government does not provide scholarships to calves,so all the money must come from the university's limited fund (whose total money is F, 0 <= F <= 2,000,000,000).
Worse still, Moo U only has classrooms for an odd number N (1 <= N <= 19,999) of the C (N <= C <= 100,000) calves who have applied.Bessie wants to admit exactly N calves in order to maximize educational opportunity. She still wants the median CSAT score of the admitted calves to be as high as possible.
Recall that the median of a set of integers whose size is odd is the middle value when they are sorted. For example, the median of the set {3, 8, 9, 7, 5} is 7, as there are exactly two values above 7 and exactly two values below it.
Given the score and required financial aid for each calf that applies, the total number of calves to accept, and the total amount of money Bessie has for financial aid, determine the maximum median score Bessie can obtain by carefully admitting an optimal set of calves.
Input
* Lines 2..C+1: Two space-separated integers per line. The first is the calf's CSAT score; the second integer is the required amount of financial aid the calf needs
Output
Sample Input
3 5 70
30 25
50 21
20 20
5 18
35 30
Sample Output
35
Hint
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<queue>
#include<functional>
using namespace std;
const int N_MAX = +;
struct cow {
int score;
int need_money;
bool operator<(const cow&b)const{
return score<b.score ;
}
};
cow cows[N_MAX];
int lower_money[N_MAX];
int upper_money[N_MAX];
int main() {
//cout << 0x3f3f3f3f <<" "<<INT_MAX<<endl;
int N, C, F;
scanf("%d%d%d",&N,&C,&F);
for (int i = ;i < C;i++)
scanf("%d%d",&cows[i].score,&cows[i].need_money);
sort(cows,cows+C);//按成绩从低到高排
priority_queue<int>que;
unsigned int half = N / ,total=;
for (int i = ;i < C;i++) {//对于每一头牛,计算成绩比他低的牛的资金总和的最小值
lower_money[i] = ( que.size() == half ? total : 0x3f3f3f3f);
que.push(cows[i].need_money);
total += cows[i].need_money;
if (que.size() > half) {
total -= que.top();
que.pop();
}
} priority_queue<int>q;
total = ;
for (int i = C - ;i >= ;i--) {
upper_money[i] = (q.size() == half ? total : 0x3f3f3f3f);
q.push(cows[i].need_money);
total += cows[i].need_money;
if (q.size() > half) {
total -= q.top();
q.pop();
}
}
int grade=-;
for (int i = C-;i>=;i--) {
if (upper_money[i] + lower_money[i] + cows[i].need_money <= F) { grade = cows[i].score; break; }
}
if (grade>)cout << grade << endl;
else cout << grade << endl;
return ;
}
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