This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#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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