POJ 2010 Moo University - Financial Aid （优先队列 | 二分查找）

Bessie noted that although humans have many universities they can attend, cows have none. To remedy this problem, she and her fellow cows formed a new university called The University of Wisconsin-Farmside,"Moo U" for short.

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

• Line 1: Three space-separated integers N, C, and F

• 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

• Line 1: A single integer, the maximum median score that Bessie can achieve. If there is insufficient money to admit N calves,output -1.
  Sample Input
  3 5 70
  30 25
  50 21
  20 20
  5 18
  35 30
  Sample Output
  35
  Hint
  Sample output:If Bessie accepts the calves with CSAT scores of 5, 35, and 50, the median is 35. The total financial aid required is 18 + 30 + 21 = 69 <= 70.

题意：

奶牛大学：奶大招生，从C头奶牛中招收N头。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。

题解：

一定要弄懂题意。然后这道题有两种解法，第一种，可以用优先队列维护前k个（和后k个）数的花费最小值，然后从按分数排名后往前找。第二种，二分查找。
优先队列：

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn=1e5+5;
int n,c,f;
typedef long long LL;
LL low[maxn],high[maxn];
struct node
{
    int id,cost;
    bool operator <(const node& a) const
    {
        return id<a.id;
    }
}a[maxn];
int main()
{
    ios::sync_with_stdio(false);
    priority_queue<int,vector<int>,less<int> >que;
    cin>>n>>c>>f;
    for(int i=0;i<c;i++)
        cin>>a[i].id>>a[i].cost;
    sort(a,a+c);
    n>>=1;
    LL sum=0;
    for(int i=0;i<n;i++)
    {
        que.push(a[i].cost);
        sum+=a[i].cost;
    }
    for(int i=n;i<=c-n-1;i++)
    {
        low[i]=sum;
        if(a[i].cost<que.top())
        {
            sum-=que.top();
            que.pop();
            que.push(a[i].cost);
            sum+=a[i].cost;
        }
    }
    sum=0;
    while(que.size()) que.pop();
    for(int i=c-1;i>c-n-1;i--)
    {
        que.push(a[i].cost);
        sum+=a[i].cost;
    }
    for(int i=c-n-1;i>=n;i--)
    {
        high[i]=sum;
        if(a[i].cost<que.top())
        {
            sum-=que.top();
            que.pop();
            que.push(a[i].cost);
            sum+=a[i].cost;
        }
    }
    int ans=-1;
    for(int i=c-n-1;i>=n;i--)
    {
        if(low[i]+a[i].cost+high[i]<=f)
        {
            ans=a[i].id;
            break;
        }
    }
    cout<<ans<<endl;
    return 0;
}

二分查找

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1e5+5;
struct cow
{
    int rank,id,cost;
}a[maxn],s[maxn];//a数组按分数排序，s数组按花费排序
int n,c,f;
bool cmp1(cow a,cow b)
{
    return a.id<b.id;
}
bool cmp2(cow a,cow b)
{
    return a.cost<b.cost;
}
int main()
{
    //ios::sync_with_stdio(false);
    scanf("%d%d%d",&n,&c,&f);
    n>>=1;
    for(int i=0;i<c;i++)
        scanf("%d%d",&a[i].id,&a[i].cost);
    sort(a,a+c,cmp1);
    for(int i=0;i<c;i++)
        a[i].rank=i;
    memcpy(s,a,sizeof(cow)*c);
    sort(s,s+c,cmp2);
    int lb=0,ub=c-1;
    int ans;
    while(ub-lb>1)
    {
        int mid=(lb+ub)>>1;
        int l=0,r=0,sum=a[mid].cost;
        for(int i=0;i<c;i++)//排序是为了先选取花费小的
        {
            if(l<n&&s[i].rank<mid&&sum+s[i].cost<=f)
            {
                l++;
                sum+=s[i].cost;
            }
            else if(r<n&&s[i].rank>mid&&sum+s[i].cost<=f)
            {
                r++;
                sum+=s[i].cost;
            }
        }
        if(l<n&&r<n)
        {
            ans=-1;
            break;
        }
        else if(l<n)
        {
            lb=mid;
        }
        else if(r<n)
        {
            ub=mid;
        }
        else
        {
            ans=a[mid].id;
            lb=mid;
        }
    }
    printf("%d\n",ans);
    return 0;
}