题解 多重影分身之术

传送门

发现可以二分答案
于是左端点一定要先把左边的走完
于是每个点两种可能，先向左走或先向右走
\(O(n)\) 扫一遍check即可

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 300010
#define ll 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, m;
int x[N], y[N];

namespace force{
	int bel[N], ans=INF;
	bool check(int t) {
		for (int i=1; i<=n; ++i) {
			int l=x[i], r=x[i];
			for (int j=1; j<=m; ++j) if (bel[j]==i) {
				l=min(l, y[j]); r=max(r, y[j]);
			}
			int dlt1=x[i]-l, dlt2=r-x[i];
			if (min(2*dlt1+dlt2, dlt2+2*dlt1)>t) return 0;
		}
		return 1;
	}
	void dfs(int u) {
		if (u>m) {
			for (int i=1; ; ++i) {
				if (check(i)) {ans=min(ans, i); break;}
			}
			return ;
		}
		for (int i=1; i<=n; ++i) {
			bel[u]=i;
			dfs(u+1);
		}
	}
	void solve() {
		dfs(1);
		printf("%d\n", ans);
		exit(0);
	}
}

namespace task1{
	bool check(int t) {
		int pos=1;
		for (int i=1; i<=n; ++i) {
			int now=t;
			if (pos<=m && y[pos]<x[i]) {
				if (now<x[i]-y[pos]) return 0;
				else now-=2*(x[i]-y[pos]);
				while (pos<=m && y[pos]<x[i]) ++pos;
			}
			while (pos<=m && y[pos]-x[i]<=now) ++pos;
		}
		return pos>m;
	}
	void solve() {
		int l=1, r=1e9+10, mid;
		while (l<=r) {
			mid=(l+r)>>1;
			if (!check(mid)) l=mid+1;
			else r=mid-1;
		}
		printf("%d\n", l);
		exit(0);
	}
}

namespace task2{
	bool check(int t) {
		int pos=1;
		for (int i=1; i<=n; ++i) {
			int now=t;
			if (pos<=m && y[pos]<x[i]) {
				if (now<x[i]-y[pos]) return 0;
				int dlt=x[i]-y[pos], reach;
				if ((now-dlt)/2 >= now-2*dlt) {
					reach=x[i]+(now-dlt)/2;
					while (pos<=m && y[pos]<=reach) ++pos;
				}
				else {
					reach=x[i]+(now-2*dlt);
					while (pos<=m && y[pos]<=reach) ++pos;
				}
			}
			else while (pos<=m && y[pos]-x[i]<=now) ++pos;
		}
		return pos>m;
	}
	void solve() {
		int l=1, r=1e9+10, mid;
		while (l<=r) {
			mid=(l+r)>>1;
			if (!check(mid)) l=mid+1;
			else r=mid-1;
		}
		printf("%d\n", l);
		exit(0);
	}
}

namespace task3{
	bool check(ll t) {
		int pos=1;
		for (int i=1; i<=n&&pos<=m; ++i) {
			ll now=t;
			if (pos<=m && y[pos]<x[i]) {
				if (now<x[i]-y[pos]) return 0;
				ll dlt=x[i]-y[pos], reach=x[i]+max((now-dlt)/2, now-2*dlt);
				while (pos<=m && y[pos]<=reach) ++pos;
			}
			else while (pos<=m && y[pos]-x[i]<=now) ++pos;
		}
		return pos>m;
	}
	void solve() {
		ll l=1, r=1e10+10, mid;
		while (l<=r) {
			mid=(l+r)>>1;
			if (!check(mid)) l=mid+1;
			else r=mid-1;
		}
		printf("%lld\n", l);
		exit(0);
	}
}

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

	n=read(); m=read();
	for (int i=1; i<=n; ++i) x[i]=read();
	for (int i=1; i<=m; ++i) y[i]=read();
	sort(x+1, x+n+1); sort(y+1, y+m+1);
	task3::solve();
	// force::solve();

	return 0;
}