Submission #1126506

#TimeUsernameProblemLanguageResultExecution timeMemory
1126506hahahahaSprinklers (CEOI24_sprinklers)C++20
100 / 100
441 ms4288 KiB
#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]) { 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 = 0, 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; }
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