AGC018C Coins

atcoder
(老版atcoder和新版的访问速度不是一个级别的（划掉)

这个题一个很关键的点：只考虑(x,y),不考虑(z)

我们假设(i)选择(A_i),(j)选择(B_j)比两者交换选择时更优,则有(A_i+B_j>A_j+B_i),移项得(A_i-B_i>A_j-B_j).做到这里一个显然的贪心就出来了：我们将所有的人按照(A_i-B_i)从大到小排序,前(x)个人选择(A_i),剩下的(y)个人选择(B_i)即为最优决策

那么现在再来考虑(z)的情况,首先继续按照(A_i-B_i)排序,那么同样不会存在排在前面的选择(B_i)同时排在后面的选择了(A_i)的情况,因此必然存在一个分界点使得分界点之前的一段不会出现(B_i),分界点之后的一段不会出现(A_i).再考虑(C_i)就会发现这两段分别是与最开始相同的问题,我们对每个子问题开一个堆,同时在枚举分界点的过程中维护当前所选择的(A_i)或者是(B_i)即可,具体过程可以参照下面的代码

#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
const int N=10000;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define fir first
#define sec second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define maxd 998244353
#define eps 1e-8
struct pnode{int a,b,c;}p[300100];
bool operator <(pnode p,pnode q) {return p.a-p.b>q.a-q.b;}
struct node{int a,b;};
bool operator <(node p,node q) {return p.a-p.b>q.a-q.b;}
priority_queue<node> q;
int n,na,nb,nc;
ll f[300300],g[300300];

int read()
{
    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

int main()
{
	na=read();nb=read();nc=read();n=na+nb+nc;
	rep(i,1,n) 
	{
		p[i].a=read();p[i].b=read();p[i].c=read();
	}
	sort(p+1,p+1+n);
	//rep(i,1,n) cout << p[i].a << " " << p[i].b << " " << p[i].c << endl;
	ll sum=0;
	rep(i,1,na) 
	{
		node tmp;
		tmp.a=p[i].a;tmp.b=p[i].c;
		q.push(tmp);sum+=p[i].a;
	}
	f[na]=sum;
	rep(i,na+1,na+nc)
	{
		node now=q.top();
		if (now.a-now.b<p[i].a-p[i].c)
		{
			sum-=now.a;sum+=now.b;
			q.pop();
			node tmp;
			tmp.a=p[i].a;tmp.b=p[i].c;sum+=p[i].a;
			q.push(tmp);
		}
		else sum+=p[i].c;
		f[i]=sum;
	}
	sum=0;
	while (!q.empty()) q.pop();
	//rep(i,1,n) cout << f[i] << " ";cout << endl;
	per(i,n,n-nb+1)
	{
		node tmp;
		tmp.a=p[i].b;tmp.b=p[i].c;
		q.push(tmp);sum+=p[i].b;
	}
	g[n-nb+1]=sum;
	per(i,n-nb,na+1)
	{
		node now=q.top();
		//cout << now.a << " " << now.b << endl;
		if (now.a-now.b<p[i].b-p[i].c)
		{
			sum-=now.a;sum+=now.b;
			q.pop();
			node tmp;
			tmp.a=p[i].b;tmp.b=p[i].c;sum+=p[i].b;
			q.push(tmp);
		}
		else sum+=p[i].c;
		g[i]=sum;
	}
	ll ans=0;
	//rep(i,1,n) cout << g[i] << " ";cout << endl;
	rep(i,na,n-nb+1) ans=max(ans,f[i]+g[i+1]);
	printf("%lld",ans);
	return 0;
}