제출 #1125920

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1125920		pan	Sprinklers (CEOI24_sprinklers)	C++20	26 / 100	409 ms	4288 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define show4(x,y,z, a) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << " " << #a << " is " << a << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
using namespace std;
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;

ll n, m;
vector<ll> xx, yy;
string solve(ll x)
{
	vector<ll> dp(n + 5, 0);
	vector<ll> bt(n+5, 0);
	for (ll i=1; i<=n; ++i)
	{
		dp[i] = dp[i-1], bt[i] = 0;
		// take L
		//ll from = lb(yy.begin()+1, yy.begin() + m  +1, xx[i] - x)-yy.begin();
		ll to = ub(yy.begin()+1, yy.begin() + m  +1, xx[i])-yy.begin() - 1;
		if (yy[dp[i-1]+1] >= xx[i] - x && yy[dp[i-1]+1] <= xx[i]) if (to >= dp[i]) dp[i] = to, bt[i] = 0;
		// take R
		//from = lb(yy.begin()+1, yy.begin() + m  +1, xx[i])-yy.begin();
		to = ub(yy.begin()+1, yy.begin() + m  +1, xx[i] + x)-yy.begin() - 1;
		if (yy[dp[i-1]+1] >= xx[i] && yy[dp[i-1]+1] <= xx[i] + x) if (to >= dp[i]) dp[i] = to, bt[i] = 1;
		// take RL
		if (i > 1 && yy[dp[i-2]+1] >= xx[i] - x && yy[dp[i-2]+1] <= xx[i])
		{
			//from = lb(yy.begin()+1, yy.begin() + m  +1, xx[i] - x)-yy.begin();
			to = ub(yy.begin()+1, yy.begin() + m  +1, xx[i])-yy.begin() - 1;
			if (to==m) {dp[i] = m, bt[i] = 2; continue;}
			if (yy[to+ 1] >= xx[i-1] && yy[to + 1] <= xx[i] + x)
			{
				to = ub(yy.begin()+1, yy.begin() + m  +1, xx[i-1] + x)-yy.begin() - 1;
				if (to >= dp[i]) dp[i] = to, bt[i] = 2;
			}
		}
	}
	if (dp[n] < m) return "";
	string ans;
	ll i = n;
	while (i > 0)
	{
		if (bt[i] == 0) ans += 'L', i--;
		else if (bt[i]==1) ans += 'R', i--;
		else ans += "LR", i-=2;
	}
	reverse(all(ans));
	return ans;
	
}


int main()
{
	cin >> n >> m;
	xx.resize(n+5), yy.resize(m+5);
	for (ll i=1; i<=n; ++i) cin >> xx[i];
	for (ll i=1; i<=m; ++i) cin >> yy[i];
	ll lo = 1, hi = 1e9;
	while (lo!=hi)
	{
		ll mid = (lo+hi) >> 1;
		string ret = solve(mid);
		if (ret == "") lo = mid+1;
		else hi = mid;
	}
	if (solve(lo)=="") cout << -1 << endl;
	else cout << lo << endl << solve(lo) << endl;
	return 0;
}

• Subtask 0: 0 / 0
• Subtask 1: 3 / 3
• Subtask 2: 6 / 6
• Subtask 3: 17 / 17
• Subtask 4: 0 / 27
• Subtask 5: 0 / 47