# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
868939 | noobcodur | Collecting Stamps 3 (JOI20_ho_t3) | C11 | 0 ms | 0 KiB |
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>
#include<ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
// #define _GLIBCXX_DEBUG 1
// #define _GLIBCXX_DEBUG_PEDANTIC 1
// #pragma GCC optimize("trapv")
// #define dbg(TXTMSG) cerr << "\n" << TXTMSG
// #define dbgv(VARN) cerr << "\n" << #VARN << " = "<< VARN << ", line: " << __LINE__ << "\n"
#define ll long long
#define ld long double
#define forn(i,j) for(ll i = 0; i < j; i++)
#define forrange(i,j,k) for(int i = j; i < k; ++i)
#define rof(i,j) rof(int i = j; i >= 0; --i)
#define pii pair<int,int>
#define vll vector<ll>
#define vi vector<int>
#define vvll vector<vll>
#define vvi vector<vi>
#define vb vector<bool>
#define pb push_back
#define p push
#define f first
#define s second
#define all(x) x.begin(), x.end()
#define eb emplace_back
#define qi queue<int>
#define qpii queue<pii>
const ll MOD = (ll)1e9+7;
void setIO(string name = ""){
ios_base::sync_with_stdio(0);
cin.tie(0);
if(!name.empty()){
freopen((name + ".in").c_str(), "r", stdin);
freopen((name + ".out").c_str(), "w", stdout);
}
}
int32_t main(){
int n, l;
cin >> n >> l;
vi x(n+1);
vi t(n+1);
x[0] = 0;
forrange(i,1,n+1){
cin >> x[i];
}
forrange(i,1,n+1){
cin >> t[i];
}
ll dp1[n+1][n+1][n+1];
ll dp2[n+1][n+1][n+1];
forn(i,n+1){
forn(j,n+1){
forn(k,n+1){
dp1[i][j][k] = MOD;
dp2[i][j][k] = MOD;
}
}
}
//1 -- joi kun ends at left
//2 -- joi kun ends at right
dp1[0][0][0] = 0;
dp2[0][0][0] = 0;
int it = 0;
forn(j,n+1){
forn(k,n+1){
dp1[n][j][k + (l - x[n] <= t[n])] = min(dp1[n][j][k + (l - x[n] <= t[n])], dp1[it][j][k] + (l - x[n]));
dp1[n][j][k + (l - x[n] + x[j] <= t[n])] = min(dp1[n][j][k + (l - x[n] + x[j] <= t[n])], dp2[it][j][k] + l - x[n] + x[j]);
}
}
for(int i = n; i > 0; --i){
forn(j,n){
forn(k,n+1){
dp1[i-1][j][k + (dp1[i][j][k] + x[i] - x[i-1] <= t[i-1])] = min(dp1[i-1][j][k + (dp1[i][j][k] + x[i] - x[i-1] <= t[i-1])], dp1[i][j][k] + x[i] - x[i-1]);
dp1[i-1][j][k + (dp2[i][j][k] + l - x[i-1] + x[j] <= t[n])] = min(dp1[i-1][j][k + (dp2[i][j][k] + l - x[i-1] + x[j] <= t[n])], dp2[i][j][k] + l - x[i-1] + x[j]);
dp2[i][j+1][k + (dp2[i][j][k] + x[j+1] - x[j] <= t[j+1])] = min(dp2[i][j+1][k + (dp2[i][j][k] + x[j+1] - x[j] <= t[j+1])], dp2[i][j][k] + x[j+1] - x[j]);
dp2[i][j+1][k + (dp1[i][j][k] + (l - x[i]) + x[j+1] <= t[j+1])] = min(dp2[i][j+1][k + (dp1[i][j][k] + (l - x[i]) + x[j+1] <= t[j+1])], dp1[i][j][k] + (l - x[i]) + x[j+1]);
}
}
}
ll ans = 0;
forn(i,n+1){
forn(j,n+1){
forn(k,n+1){
if((dp1[i][j][k] != MOD) || (dp2[i][j][k] != MOD)){
ans = max(ans,k);
}
}
}
}
cout << ans << endl;
}