Submission #536053

#TimeUsernameProblemLanguageResultExecution timeMemory
536053browntoadCollecting Stamps 3 (JOI20_ho_t3)C++14
15 / 100
87 ms135288 KiB
#include <bits/stdc++.h> #pragma GCC optimize ("Ofast", "unroll-loops") using namespace std; #define ll long long #define int ll #define FOR(i,a,b) for (int i = (a); i<(b); i++) #define REP(i,n) FOR(i,0,n) #define REP1(i,n) FOR(i,1,n+1) #define RREP(i,n) for (int i=(n)-1; i>=0; i--) #define RREP1(i,n) for (int i=(n); i>=1; i--) #define f first #define s second #define pb push_back #define ALL(x) x.begin(),x.end() #define SZ(x) (int)(x.size()) #define SQ(x) (x)*(x) #define pii pair<int, int> #define pdd pair<double ,double> #define pcc pair<char, char> #define endl '\n' //#define TOAD #ifdef TOAD #define bug(x) cerr<<__LINE__<<": "<<#x<<" is "<<x<<endl #define IOS() #else #define bug(...) #define IOS() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0) #endif const ll inf = 1ll<<60; const int iinf=2147483647; const ll mod = 1e9+7; const ll maxn=5005; const double PI=acos(-1); ll pw(ll x, ll p, ll m=mod){ ll ret=1; while (p>0){ if (p&1){ ret*=x; ret%=m; } x*=x; x%=m; p>>=1; } return ret; } ll inv(ll a, ll m=mod){ return pw(a,m-2); } //======================================================================================= int dp[205][205][205][2]; signed main (){ IOS(); int n, l; cin>>n>>l; vector<int> vc(n+2), ti(n+2); REP1(i,n){ cin>>ti[i]; } REP1(i,n){ cin>>vc[i]; } ti[n+1]=l; //cout<<ti[6]<<endl; REP(i,205){ REP(j,205){ REP(t, 205){ REP(k, 2){ dp[i][j][t][k]=-inf; } } } } dp[0][n+1][0][0]=dp[0][n+1][0][1]=0; //dp[0][n][l-vc[n]][1]=((l-vc[n])<=ti[n]); //dp[1][n+1][vc[1]][0]=((vc[1])<=ti[1]); int ans=0; REP(i,n+1){ for (int j=n+1; j>i; j--){ if (i==0&&j==n+1) continue; if (i>0) { REP(t, 205){ int a=t-(ti[i]-ti[i-1]); if (a>=0) dp[i][j][t][0]=max(dp[i][j][t][0], dp[i-1][j][a][0]); int b=t-(ti[i]+l-ti[j]); if (b>=0) dp[i][j][t][0]=max(dp[i][j][t][0], dp[i-1][j][b][1]); dp[i][j][t][0]+=(vc[i]>=t); } } if (j<=n) { REP(t, 205){ int a=t-(ti[j+1]-ti[j]); if (a>=0) dp[i][j][t][1]=max(dp[i][j][t][1], dp[i][j+1][a][1]); int b=t-(ti[i]+l-ti[j]); if (b>=0) dp[i][j][t][1]=max(dp[i][j][t][1], dp[i][j+1][b][0]); dp[i][j][t][1]+=(vc[j]>=t); } } REP(t, 205){ ans=max({ans, dp[i][j][t][0], dp[i][j][t][1]}); } } } //cout<<dp[0][5][3][1]<<endl; //cout<<dp[0][4][7][1]<<endl; cout<<ans<<endl; }
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