题解 开挂

传送门

可以证明一个结论：
若从后向前扫，每个重复的数变为它后面第一个没有出现过的数一定更优
于是可以用并查集找这样的未出现的数
复杂度 \(O(nlogn)\)

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1000010
#define ll long long
#define ull unsigned long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;
ll a[N], b[N];

namespace force{
	ll sta[N], top;
	bool vis[N];
	ull ans;
	void solve() {
		sort(a+1, a+n+1);
		sort(b+1, b+n+1);
		for (int i=n; i; --i) {
			if (!vis[a[i]]) {vis[a[i]]=1; sta[++top]=0;}
			else {
				int j;
				for (j=a[i]; vis[j]; ++j) ;
				sta[++top]=j-a[i];
				vis[j]=1;
			}
		}
		sort(sta+1, sta+top+1, [](ll a, ll b){return a>b;});
		// cout<<"sta: "; for (int i=1; i<=top; ++i) cout<<sta[i]<<' '; cout<<endl;
		for (int i=1; i<=n; ++i) ans+=1llu*sta[i]*b[i];
		printf("%llu\n", ans);
		exit(0);
	}
}

namespace task1{
	ll sta[N];
	int dsu[N], nxt[N], tot, top;
	ull ans;
	unordered_map<ll, int> mp;
	inline int find(int p) {return dsu[p]==p?p:dsu[p]=find(dsu[p]);}
	void solve() {
		sort(a+1, a+n+1);
		sort(b+1, b+n+1);
		for (int i=n,f,tem; i; --i) {
			if (mp.find(a[i])==mp.end()) {
				mp[a[i]]=++tot;
				dsu[tot]=tot;
				sta[++top]=0;
				if (mp.find(a[i]+1)==mp.end()) nxt[tot]=a[i]+1;
				else dsu[tot]=find(mp[a[i]+1]);
			}
			else {
				f=find(mp[a[i]]);
				tem=nxt[f];
				mp[tem]=++tot;
				dsu[tot]=f;
				sta[++top]=tem-a[i];
				if (mp.find(tem+1)==mp.end()) nxt[f]=tem+1;
				else dsu[f]=find(mp[tem+1]);
			}
		}
		sort(sta+1, sta+top+1, [](ll a, ll b){return a>b;});
		// cout<<"sta: "; for (int i=1; i<=top; ++i) cout<<sta[i]<<' '; cout<<endl;
		for (int i=1; i<=n; ++i) ans+=1llu*sta[i]*b[i];
		printf("%llu\n", ans);
		exit(0);
	}
}

signed main()
{
	freopen("openhook.in", "r", stdin);
	freopen("openhook.out", "w", stdout);

	n=read();
	for (int i=1; i<=n; ++i) a[i]=read();
	for (int i=1; i<=n; ++i) b[i]=read();
	// force::solve();
	task1::solve();

	return 0;
}