Submission #1092411

#	Time	Username	Problem	Language	Result	Execution time	Memory
1092411		vjudge1	Pinball (JOI14_pinball)	C++17	11 / 100	182 ms	524288 KiB

pinball

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#pragma GCC target("popcnt")
using namespace std;
 
using ll = long long;
using ull = unsigned long long;
using lld = long double;
using ii = pair<int,int>;
using pll = pair<ll, ll>;
 
using vi = vector<int>;
using vll = vector<ll>;
using vii = vector<ii>;
using vpll = vector<pll>;
using vlld = vector<lld>;
 
// #define endl '\n'
#define all(x) x.begin(),x.end()
#define lsb(x) x&(-x)
#define gcd(a,b) __gcd(a,b)
#define sz(x) (int)x.size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fls cout.flush()
 
#define fore(i,l,r) for(auto i=l;i<r;i++)
#define fo(i,n) fore(i,0,n)
#define forex(i,r,l) for(auto i=r; i>=l;i--)
#define ffo(i,n) forex(i,n-1,0)
 
bool cmin(ll &a, ll b) { if(b<a){a=b;return 1;}return 0; }
bool cmax(ll &a, ll b) { if(b>a){a=b;return 1;}return 0; }
void valid(ll in) { cout<<((in)?"YES\n":"NO\n"); }
ll lcm(ll a, ll b) { return (a/gcd(a,b))*b; }
ll gauss(ll n) { return (n*(n+1))/2; }

const int N = 204;
ll dp[N][3*N][3*N];
/*
dp[i][c1][cn]
estando en la fila i cual es la minima suma de costos para que:
la ficha 1 este en la posicion rd[c1] y la ficha n este en la posicion rd[c2]
*/
struct device { int a, b, c, d; };
void test_case() {
    int n, m;
    cin >> n >> m;
    vector<device> a(n+1);
    deque<ll> rd;
    fore (i, 1, n+1) {
        cin >> a[i].a >> a[i].b >> a[i].c >> a[i].d;
        rd.pb(a[i].a);
        rd.pb(a[i].b);
        rd.pb(a[i].c);
    }
    rd.pb(1);
    rd.pb(m);
    sort(all(rd));
    fo(i, sz(rd)-1) if(rd[i+1] == rd[i]) rd[i] = 0;
    sort(all(rd));
    while(rd[0]==0) rd.pop_front();
    fo (i, n+1) fo(i2, 3*n+3) fo(i3, 3*n+3) dp[i][i2][i3] = 1e18;
    dp[0][0][sz(rd)-1]=0;
    // for (int v: rd) cout << v << " ";
    // cout << '\n';
    // return;
    fore (i, 1, n+1) {
        // no hacer nada
        fo (i2, sz(rd)) fo(i3, sz(rd)) dp[i][i2][i3] = dp[i-1][i2][i3];
        // usar este y que se modifique al menos la posicion del 1
        int p = lower_bound(all(rd), a[i].c)-rd.begin();
        fo (i2, sz(rd)) {
            fo (i3, sz(rd)) {
                int p2 = (a[i].a<=rd[i2] && rd[i2]<=a[i].b ? p : i2);
                int p3 = (a[i].a<=rd[i3] && rd[i3]<=a[i].b ? p : i3);
                dp[i][p2][p3] = min(dp[i][p2][p3], dp[i-1][i2][i3] + a[i].d);
            }
        }
    }
    ll ans = 1e18;
    fo (i, sz(rd)) ans=min(ans, dp[n][i][i]);
    if (ans == 1e18) cout << "-1\n";
    else cout << ans << '\n';
}

int main() {
    int tt = 1;
    // cin >> tt;
    while(tt--) test_case();
}

Compilation message (stderr)

pinball.cpp:4: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
    4 | #pragma GCC optimization ("O3")
      | 
pinball.cpp:5: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
    5 | #pragma GCC optimization ("unroll-loops")
      |

• Subtask 1: 11 / 11
• Subtask 2: 0 / 18
• Subtask 3: 0 / 22
• Subtask 4: 0 / 49