이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define debug(x) cerr << #x << " : " << x << '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
typedef string str;
typedef pair<ll, ll> pll;
const int N = 2e2 + 10;
const ll Inf = 1e15;
const int Mx = 1e9 + 1;
ll n, L;
ll x[N], t[N], dp[N][N][N][2];
ll Dis(ll i, ll j){
return min(abs(i - j), L - abs(i - j));
}
int main(){
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
cin >> n >> L;
for(int i = 0; i < n; i++)
cin >> x[i];
for(int i = 0; i < n; i++)
cin >> t[i];
ll ans = 0;
memset(dp, 31, sizeof dp);
dp[0][0][0][0] = 0;
dp[0][0][0][1] = 0;
ll T, I, J;
for(int i = 0; i < n; i++){
I = (i ? x[i - 1] : 0);
for(int j = 0; i + j < n; j++){
J = (j ? x[n - j] : 0);
for(int k = 0; k <= i + j; k++){
T = dp[i][j][k][0] + Dis(I, x[i]);
dp[i + 1][j][k + (T <= t[i])][0] = min(dp[i + 1][j][k + (T <= t[i])][0], T);
T = dp[i][j][k][0] + Dis(I, x[n - j - 1]);
dp[i][j + 1][k + (T <= t[n - j - 1])][1] = min(dp[i][j + 1][k + (T <= t[n - j - 1])][1], T);
T = dp[i][j][k][1] + Dis(J, x[i]);
dp[i + 1][j][k + (T <= t[i])][0] = min(dp[i + 1][j][k + (T <= t[i])][0], T);
T = dp[i][j][k][1] + Dis(J, x[n - j - 1]);
dp[i][j + 1][k + (T <= t[n - j - 1])][1] = min(dp[i][j + 1][k + (T <= t[n - j - 1])][1], T);
}
}
}
for(int i = 0; i <= n; i++)
for(int k = 0; k <= n; k++)
if(min(dp[i][n - i][k][0], dp[i][n - i][k][1]) < Inf) ans = max(ans, (ll) k);
cout << ans << '\n';
return 0;
}
/*
10 87
9 23 33 38 42 44 45 62 67 78
15 91 7 27 31 53 12 91 89 4
*/
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