Submission #577276

#TimeUsernameProblemLanguageResultExecution timeMemory
577276MohammadAghilCollecting Stamps 3 (JOI20_ho_t3)C++17
0 / 100
1 ms980 KiB
#include <bits/stdc++.h> // #pragma GCC optimize ("Ofast,unroll-loops") // #pragma GCC target ("avx2") using namespace std; typedef long long ll; typedef pair<ll, int> pp; #define er(args ...) cerr << __LINE__ << ": ", err(new istringstream(string(#args)), args), cerr << endl #define per(i,r,l) for(int i = (r); i >= (l); i--) #define rep(i,l,r) for(int i = (l); i < (r); i++) #define all(x) x.begin(), x.end() #define sz(x) (int)(x).size() #define pb push_back #define ss second #define ff first void err(istringstream *iss){}template<typename T,typename ...Args> void err(istringstream *iss,const T &_val, const Args&...args){string _name;*iss>>_name;if(_name.back()==',')_name.pop_back();cerr<<_name<<" = "<<_val<<", ",err(iss,args...);} template<typename T> bool chkmin(T &a, T b){return (b < a) ? a = b, 1 : 0;} template<typename T> bool chkmax(T &a, T b){return (b > a) ? a = b, 1 : 0;} mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); const ll mod = 1e9 + 7, maxn = 2e2 + 5, lg = 22, inf = ll(1e9) + 5; ll pw(ll a,ll b,ll md=mod){if(!b)return 1;ll k=pw(a,b>>1ll);return k*k%md*(b&1ll?a:1)%md;} ll dp[maxn][maxn][maxn<<1][2]; int main(){ cin.tie(0) -> sync_with_stdio(0); int n, l; cin >> n >> l; vector<int> x(n+1), t(n+1); x[0] = 0; rep(i,1,n+1) cin >> x[i]; rep(i,1,n+1) cin >> t[i]; vector<int> xn(n+1), tn(n+1); rep(i,1,n+1) xn[i] = l - x[n+1-i], tn[i] = t[n+1-i]; int ans = 0; rep(i,0,n+1) rep(j,0,n+1) rep(k,0,(n<<1)+1) rep(t,0,2) dp[i][j][k][t] = inf; dp[0][0][0][0] = dp[0][0][0][1] = 0; rep(i,0,n+1){ rep(j,0,n+1){ rep(k,0,(n<<1)+1){ rep(t,0,2) if(dp[i][j][k][t] < inf) ans = max(ans, min(n, k)); // er(i, j, k); if(i < n){ ll tm = dp[i][j][k][0] + xn[i+1] - xn[i]; chkmin(dp[i+1][j][k + (tm <= tn[i+1])][0], tm); tm = dp[i][j][k][1] + x[j] + xn[i+1]; chkmin(dp[i+1][j][k + (tm <= tn[i+1])][0], tm); } if(j < n){ ll tm = dp[i][j][k][0] + xn[i] + x[j+1]; chkmin(dp[i][j+1][k + (tm <= t[j+1])][1], tm); tm = dp[i][j][k][1] + x[j+1] - x[j]; chkmin(dp[i][j+1][k + (tm <= t[j+1])][1], tm); } } } } cout << ans << '\n'; return 0; }
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