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

题目大意：有C头牛，每头牛都有相应的分数和需求，要求在这C头牛中选出N头，使得这N头牛中的分数的中位数达到最大，且需求之和小于等于F

解题思路：先按成绩排序
再用两个数组保留最小需求之和
left数组保留第i个位置左边的 N/2个最小需求之和
right数组保留第i个位置右边的 N/2个最小需求之和
如何保留最小的需求之和呢，扫描两遍（左右），用优先队列保留N / 2个最小需求
最后只需要判断一下 left[i] + right[i] + 第i头牛的需求 <= F就可以了

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define maxn 100010
typedef long long ll;

struct Cows{
    ll score, aid;
}cow[maxn];

int N, C;
ll F, left[maxn], right[maxn];

bool cmp(const Cows a, const Cows b) {
    return a.score < b.score;
}

void solve() {

    sort(cow, cow + C, cmp);
    for(int i = 0; i < C; i++)
        left[i] = right[i] = 0;
    priority_queue<ll> pq;

    ll sum1 = 0;
    for(int i = 0; i < C; i++) {

        if(i < N / 2) {
            pq.push(cow[i].aid);
            sum1 += cow[i].aid;
            continue;
        }
        left[i] = sum1;
        if(cow[i].aid >= pq.top())
            continue;
        sum1 -= pq.top();
        pq.pop();
        sum1 += cow[i].aid;
        pq.push(cow[i].aid);
    }

    while(!pq.empty())
        pq.pop();

    sum1 = 0;
    for(int i = C - 1; i >= 0; i--) {
        if(i > C - 1 - N / 2) {
            sum1 += cow[i].aid;
            pq.push(cow[i].aid);
            continue;
        }
        right[i] = sum1;
        if(cow[i].aid >= pq.top())
            continue;
        sum1 -= pq.top();
        pq.pop();
        sum1 += cow[i].aid;
        pq.push(cow[i].aid);
    }

    ll ans = -1;
    for(int i = N / 2; i <= C - 1 - N / 2; i++)
        if(left[i] + right[i] + cow[i].aid <= F) 
            ans = cow[i].score;
    printf("%lld\n", ans);
}

int main(){
    while(scanf("%d%d%lld",&N, &C, &F) != EOF) {
        for(int i = 0; i < C; i++)
            scanf("%lld%lld", &cow[i].score, &cow[i].aid);
        solve();
    }
    return 0;
}