#include <bits/stdc++.h>
using namespace std;
template<class T, class U>
void ckmin(T &a, U b)
{
if (a > b) a = b;
}
template<class T, class U>
void ckmax(T &a, U b)
{
if (a < b) a = b;
}
#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
const int MAXN = 213;
const int INF = 1000000007;
const ll LLINF = 2696969696969696969ll;
int N;
ll L;
pll arr[MAXN];
ll dp[MAXN][MAXN][2][MAXN];
int ans;
int32_t main()
{
cout << fixed << setprecision(12);
cerr << fixed << setprecision(4);
ios_base::sync_with_stdio(false); cin.tie(0);
cin >> N >> L;
FOR(i, 0, N)
{
cin >> arr[i].fi;
}
FOR(i, 0, N)
{
cin >> arr[i].se;
}
arr[N] = {L, 0};
//dp[how many left we've been][how many right we've been][where are we?][# collected] = time
FOR(i, 0, N + 1)
{
FOR(j, 0, N + 1)
{
FOR(k, 0, 2)
{
FOR(m, 0, N + 1)
{
dp[i][j][k][m] = LLINF;
}
}
}
}
dp[0][0][1][0] = 0;
FOR(i, 0, N + 1)
{
FOR(j, 0, N + 1 - i)
{
FOR(k, 0, 2)
{
FOR(m, 0, N + 1)
{
if (dp[i][j][k][m] >= LLINF) continue;
ckmax(ans, m);
if (i + j == N) continue;
//head left.
ll nt; int nm;
if (k) //you are currently at arr[N - j].fi
{
//try to go to arr[i].fi
nt = dp[i][j][k][m] + (L - arr[N - j].fi + arr[i].fi);
nm = m + (nt <= arr[i].se);
ckmin(dp[i + 1][j][0][nm], nt);
//try to go to arr[N - j - 1].fi
nt = dp[i][j][k][m] + (arr[N - j].fi - arr[N - j - 1].fi);
nm = m + (nt <= arr[N - j - 1].se);
ckmin(dp[i][j + 1][1][nm], nt);
}
else //you are currently at arr[i - 1].fi
{
//try to go to arr[i].fi
nt = dp[i][j][k][m] + (arr[i].fi - arr[i - 1].fi);
nm = m + (nt <= arr[i].se);
ckmin(dp[i + 1][j][0][nm], nt);
//try to go to arr[N - j - 1].fi
nt = dp[i][j][k][m] + (arr[i - 1].fi + L - arr[N - j - 1].fi);
nm = m + (nt <= arr[N - j - 1].se);
ckmin(dp[i][j + 1][1][nm], nt);
}
}
}
}
}
cout << ans << '\n';
}
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
5 ms |
760 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
5 ms |
760 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
5 ms |
760 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |