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>
using namespace std;
#define pos first
#define tim second
#define int long long int
const int INF = 1e18;
int32_t main() {
int n, l;
cin>>n>>l;
vector<pair<int,int>> stamps(n+2);
stamps[0] = {0,INF};
for (int i = 1; i<=n; i++) cin>>stamps[i].pos;
for (int i = 1; i<=n; i++) cin>>stamps[i].tim;
stamps[n+1] = {l, INF};
int ldp[n+5][n+5][n+5], rdp[n+5][n+5][n+5];
// whenever we move if we are moving optimally we will cover a range L, R (taking some stamps)
// ldp[L][R][i] - min time when covered L to R and we are at left end and taken i stamps
// rdp[L][R][i] - same but we are at right end
for (int i = 0; i <= n+1; i++) for (int j = 0; j <= n+1; j++) for (int k = 0; k <= n; k++) ldp[i][j][k] = rdp[i][j][k] = INF;
ldp[0][n+1][0] = rdp[0][n+1][0] = 0;
for (int L = 0; L <= n; L++) {
for (int R = n+1; R >= L+2; R--) {
for (int k = 0; k <= n; k++) {
// go L to L+1, go from R to L+1
int tl = min(ldp[L][R][k]+stamps[L+1].pos-stamps[L].pos, rdp[L][R][k]+(l-stamps[R].pos)+stamps[L+1].pos);
// go L to R-1, go from R to R-1
int tr = min(ldp[L][R][k]+(l-stamps[R-1].pos)+stamps[L].pos, rdp[L][R][k]+stamps[R].pos-stamps[R-1].pos);
// will we be able to take R-1 or L+1 stamp? check and update
ldp[L+1][R][k+(tl<=stamps[L+1].tim)] = min(ldp[L+1][R][k+(tl<=stamps[L+1].tim)], tl);
rdp[L][R-1][k+(tr<=stamps[R-1].tim)] = min(rdp[L][R-1][k+(tr<=stamps[R-1].tim)], tr);
}
}
}
int ans = 0;
for (int i = 0; i <= n+1; i++) {
for (int j = 0; j <= n+1; j++) {
for (int k = 0; k <= n; k++) {
if (ldp[i][j][k] != INF || rdp[i][j][k] != INF) ans = max(ans, k);
}
}
}
cout<<ans;
}
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