poj2010Moo University - Financial Aid

要录取n个，那中位数左右都会有（n-1）/2个。枚举中位数第i位，左右两边预处理出（i-1）前（n-1）/2个最小数的和，（i+1）后（n-1）/2个最小数的和。

我用的是优先队列。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;
struct node
{
ll scr,mon;
}edge[100005];
ll cmp(node a,node b)
{
return a.scr<b.scr;
}
ll fr[100005],af[100005];
ll cmp2(ll a,ll b)
{
return a<b;
}
ll n,c,f;
struct nn
{
ll shu;
};
priority_queue<nn>Q;//而且本人的优先队列写的真麻烦。。。
bool operator< (nn a,nn b)
{
return a.shu<b.shu;
}
int main()
{
while(scanf("%lld %lld %lld",&n,&c,&f)!=EOF)
{
memset(fr,0,sizeof(fr));
memset(af,0,sizeof(af));
for(ll i=1;i<=c;i++)
{
scanf("%lld %lld",&edge[i].scr,&edge[i].mon);
}
sort(edge+1,edge+1+c,cmp);
ll mid=(n-1)/2;
ll sum=0;
while(!Q.empty())
{
Q.pop();
}
for(ll i=1;i<=mid;i++)
{
nn s;
s.shu=edge[i].mon;
Q.push(s);
sum+=edge[i].mon;
}
fr[mid]=sum;
for(ll i=mid+1;i<=c;i++)
{
nn t=Q.top();
if(edge[i].mon<t.shu)
{
Q.pop();
nn s;
s.shu=edge[i].mon;
Q.push(s);
sum=sum-t.shu+edge[i].mon;
}
fr[i]=sum;
}
while(!Q.empty())
{
Q.pop();
}
sum=0;
for(ll i=c;i>=c-mid+1;i--)
{
nn s;
s.shu=edge[i].mon;
Q.push(s);
sum+=edge[i].mon;
}
af[c-mid+1]=sum;
for(ll i=c-mid;i>=1;i--)
{
nn t=Q.top();
if(edge[i].mon<t.shu)
{
Q.pop();
nn s;
s.shu=edge[i].mon;
Q.push(s);
sum=sum-t.shu+edge[i].mon;
}
af[i]=sum;
}
int si=0;
for(ll i=c-mid;i>=mid+1;i--)
{
if(edge[i].mon+af[i+1]+fr[i-1]<=f)
{
printf("%lld\n",edge[i].scr);
si=1;
break;
}
}
if(si==0)printf("-1\n");
}
return 0;

}

tip:

优先队列中排的是数时，什么也不加，按从大到小排列。

priority_queue<int,vector<int>,greater<int> >q;从小到大。