# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
982694 | LOLOLO | Collecting Stamps 3 (JOI20_ho_t3) | C++17 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#define ll long long
using namespace std;
#define f first
#define s second
#define pb push_back
#define ep emplace
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define uniquev(v) sort(all(v)), (v).resize(unique(all(v)) - (v).begin())
#define mem(f,x) memset(f , x , sizeof(f))
#define sz(x) (int)(x).size()
#define __lcm(a, b) (1ll * ((a) / __gcd((a), (b))) * (b))
#define mxx *max_element
#define mnn *min_element
#define cntbit(x) __builtin_popcountll(x)
#define len(x) (int)(x.length())
const int N = 210;
ll f[N][N][N][2];
ll x[N], t[N];
void minimize(ll &a, ll b) {
if (a > b)
a = b;
}
ll n, l;
ll dis(ll a, ll b) {
return min(abs(a - b), l - abs(a - b));
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
mem(f, 0x3f);
cin >> n >> l;
for (int i = 1; i <= n; i++)
cin >> x[i];
for (int i = 1; i <= n; i++)
cin >> t[i];
x[0] = 0;
x[n + 1] = ;
t[0] = -1;
t[n + 1] = -1;
f[0][0][n + 1][0] = 0;
f[0][0][n + 1][1] = 0;
int ans = 0;
for (int c = 0; c <= n; c++) {
for (int i = 0; i <= n + 1; i++) {
for (int j = n + 1; j > i; j--) {
for (int k = 0; k < 2; k++) {
ll pos = k ? x[j] : x[i];
ll lim = k ? t[j] : t[i];
if (i) {
if (c) {
ll ti = f[c - 1][i - 1][j][0] + dis(x[i - 1], pos);
if (ti <= lim) {
minimize(f[c][i][j][k], ti);
}
}
minimize(f[c][i][j][k], f[c][i - 1][j][0] + dis(x[i - 1], pos));
}
if (j != n + 1) {
if (c) {
ll ti = f[c - 1][i][j + 1][1] + dis(x[j + 1], pos);
if (ti <= lim) {
minimize(f[c][i][j][k], ti);
}
}
minimize(f[c][i][j][k], f[c][i][j + 1][1] + dis(x[j + 1], pos));
}
if (f[c][i][j][k] <= 1e14) {
ans = max(ans, c);
}
}
}
}
}
cout << ans << '\n';
return 0;
}