[TJOI2013] 奖学金

代码：

#include<bits/stdc++.h>
using namespace std;
long long n,c,ff,ans;
long long suma[200010],sumb[200010];//前  后
struct score{ long long a,b; }s[200010];
bool cmp(score aa,score bb){return aa.a<bb.a; }
priority_queue<int>q;
int main()
{
	scanf("%lld%lld%lld",&n,&c,&ff);
	for(int i=1;i<=c;i++)
	{
		scanf("%lld%lld",&s[i].a,&s[i].b);
	}
	sort(s+1,s+c+1,cmp);
	for(int i=1;i<=n/2;i++)
	{
		q.push(s[i].b);
		suma[i]=suma[i-1]+s[i].b;
	}
	for(int i=n/2+1;i<=c;i++)
	{
		if(s[i].b<q.top())
		{
			suma[i]=suma[i-1]-q.top()+s[i].b;
			q.pop();
			q.push(s[i].b);
		}
		else suma[i]=suma[i-1];
	}
	while(q.size()) q.pop();
	for(int i=c;i>=c-n/2+1;i--)
	{
		q.push(s[i].b);
		sumb[i]=sumb[i+1]+s[i].b;
	}
	for(int i=c-n/2;i>=1;i--)
	{
		if(s[i].b<q.top())
		{
			sumb[i]=sumb[i+1]-q.top()+s[i].b;
			q.pop();
			q.push(s[i].b);
		}
		else sumb[i]=sumb[i+1];
	}
	for(int i=n/2+1;i<=c-n/2;i++)
		if(suma[i-1]+sumb[i+1]+s[i].b<=ff)
			ans=s[i].a;
	if(ans==0)
	{
		printf("-1");
		return 0;
	}
	printf("%lld",ans);
	return 0;
}

整体思路在于从中位数入手，枚举每个数作为中位数，分别算出两边奖学金

用堆（优先队列）维护两边每次最小奖学金总和，时间复杂度O(nlongn)