Submission #624016

#TimeUsernameProblemLanguageResultExecution timeMemory
624016jiahngUplifting Excursion (BOI22_vault)C++14
0 / 100
80 ms45692 KiB
#include <bits/stdc++.h> using namespace std; typedef long long ll; #define int ll typedef pair<int,int> pi; typedef vector <int> vi; typedef vector <pi> vpi; typedef pair<pi, ll> pii; typedef set <ll> si; typedef long double ld; #define f first #define s second #define mp make_pair #define FOR(i,s,e) for(int i=s;i<=int(e);++i) #define DEC(i,s,e) for(int i=s;i>=int(e);--i) #define pb push_back #define all(x) (x).begin(), (x).end() #define lbd(x, y) lower_bound(all(x), y) #define ubd(x, y) upper_bound(all(x), y) #define aFOR(i,x) for (auto i: x) #define mem(x,i) memset(x,i,sizeof x) #define fast ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define maxn 611 #define INF (ll)(1e18+10) #define MOD 998244353 typedef pair <vi, int> pvi; typedef pair <int,pi> ipi; typedef vector <pii> vpii; typedef pair <pi,pi> pipi; int dp[maxn][maxn][maxn]; int A[maxn], sm = 0; int M,L; int32_t main(){ fast; cin >> M >> L; FOR(i,0,2*M) cin >> A[i]; vector <pii> v; int ans = 0; FOR(i,M,2*M) if (sm < L){ int x = INF; if (i != M){ x = (L - sm) / (i - M); } x = min(x, A[i]); sm += x * (i-M); A[i] -= x; v.pb(pii(pi(-(i-M), x), -1)); ans += x; assert(sm <= L); if (A[i] > 0 && sm < L){ sm += i - M; A[i]--; ans++; v.pb(pii(pi(-(i-M),1),-1)); } } //cout << sm << ' ' << ans << '\n'; //FOR(i,0,2*M) cout << A[i] << ' '; //cout << '\n'; if (sm < L){ cout << "impossible"; return 0; } //assert(sm <= L + M); FOR(i,0,2*M) v.pb(pii(pi(i-M, A[i]), 1)); vector <pii> pos, neg; aFOR(i, v) if (i.f.s > 0){ if (i.f.f >= 0) pos.pb(i); else neg.pb(i); } FOR(i,0,2*M) dp[i][0][0] = -INF; dp[M][0][0] = 0; FOR(j,0,pos.size()) FOR(k,0,neg.size()) if (j > 0 || k > 0){ FOR(i,0,2*M){ dp[i][j][k] = -INF; if (j > 0){ FOR(x,0,min(pos[j-1].f.s, 2*M)) if (0 <= i + x * pos[j-1].f.f && i + x * pos[j-1].f.f <= 2*M){ dp[i][j][k] = max(dp[i][j][k], dp[i+x*pos[j-1].f.f][j-1][k] + x * pos[j-1].s); } } if (k > 0){ FOR(x,0,min(neg[k-1].f.s, 2*M)) if (0 <= i + x * neg[k-1].f.f && i + x * neg[k-1].f.f <= 2*M){ dp[i][j][k] = max(dp[i][j][k], dp[i+x*neg[k-1].f.f][j][k-1] + x * neg[k-1].s); } } } } ans += dp[sm - L + M][pos.size()][neg.size()]; if (ans < 0) cout << "impossible"; else cout << ans; }
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