이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "overtaking.h"
#include <bits/stdc++.h>
#define int long long
#define fi first
#define se second
using namespace std;
vector<signed> W, S;
vector<long long> T;
vector<vector<int>> dp;
vector<vector<vector<int>>> arr;
int L, N, X, M;
void init(signed LL, signed NN, std::vector<long long> TT, std::vector<signed> WW, signed XX, signed MM, std::vector<signed> SS)
{
ios::sync_with_stdio(false);
cin.tie(0);
L = LL; N = NN; X = XX; M = MM; S = SS;
int n = N;
for (int i = 0; i < N; i++) {
if (WW[i] <= X) {
n--;
continue;
}
T.push_back(TT[i]);
W.push_back(WW[i]);
}
N = n;
dp.resize(M, vector<int>(N));
arr.resize(M, vector<vector<int>> (N, vector<int> (3)));
vector<vector<int>> bus(N, vector<int> (M));
for (int i = 0; i < N; i++) {
arr[0][i] = {T[i], W[i], i};
bus[i][0] = T[i];
}
sort(arr[0].begin(), arr[0].end());
for (int i = 1; i < M; i++) {
int last = 0;
for (int j = 0; j < N; j++) {
int expected = arr[i - 1][j][0] + arr[i - 1][j][1] * (S[i] - S[i - 1]);
if (expected < last) arr[i][j] = {last, arr[i - 1][j][1], arr[i - 1][j][2]};
else {
last = expected;
arr[i][j] = {expected, arr[i - 1][j][1], arr[i-1][j][2]};
}
bus[arr[i-1][j][2]][i] = last;
}
sort(arr[i].begin(), arr[i].end());
}
for (int i = 0; i < N; i++) dp[M - 1][i] = bus[i][M - 1];
for (int i = M - 2; i > 0; i--) {
for (int j = 0; j < N; j++) {
int next = bus[j][i];
signed nb = arr[i].end() - upper_bound(arr[i].begin(), arr[i].end(), vector<int>({next, -1, -1}));
signed l = i, r = M - 1;
while (l <= r) {
signed mid = l + (r - l) / 2;
if (N - nb > 0 && arr[mid][N - nb - 1][0] >= next + X * (S[mid] - S[i])) r = mid - 1;
else l = mid + 1;
}
if (l == M) dp[i][j] = next + X * (S[M - 1] - S[i]);
else dp[i][j] = dp[l][arr[l][N - nb - 1][2]];
}
}
return;
}
long long arrival_time(long long Y)
{
signed l = 1, r = M - 1;
signed ans = -1;
signed nb = arr[0].end() - upper_bound(arr[0].begin(), arr[0].end(), vector<int>({Y, -1, -1}));
while (l <= r) {
signed mid = l + (r - l) / 2;
if (N - nb > 0 && arr[mid][N - nb - 1][0] >= X * S[mid] + Y) {
ans = mid;
r = mid - 1;
}
else l = mid + 1;
}
if (ans == -1) return S[M - 1] * X + Y;
else return dp[ans][arr[ans][N - nb - 1][2]];
}
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