이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")
#define F first
#define S second
#define pb push_back
#define SZ(x) (ll)(x.size())
#define all(x) x.begin(),x.end()
#define MP make_pair
typedef int ll;
typedef long double ld;
typedef pair<ll,ll> pll;
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll maxn=2e2+10, maxm=1e3+10, lg=21, mod=1e9+7, inf=1e18;
ll n,L,X[maxn],T[maxn],dp[maxn][maxn][maxn][2],it,tt,ans=0;
void mst(ll &x,ll y){x=min(x,y);}
int main(){
ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
memset(dp,69,sizeof(dp));
cin>>n>>L;
for(int i=1;i<=n;i++) cin>>X[i];
for(int i=1;i<=n;i++) cin>>T[i];
dp[0][0][0][0]=dp[0][0][1][0]=0;
for(int b=0;b<=n;b++)for(int f=0;b+f<=n;f++)for(int k=0;k<=b+f;k++){
it=dp[b][f][k][0];
if(it<mod){ ans=max(ans,k);
tt=it+X[f+1]-X[f];
mst(dp[b][f+1][k+(tt<=T[f+1])][0],tt);
tt=it+X[f]+L-X[n-b];
mst(dp[b+1][f][k+(tt<=T[n-b])][1],tt);
}
it=dp[b][f][k][1];
if(it<mod){ ans=max(ans,k);
tt=it+X[n-b+1]-X[n-b];
mst(dp[b+1][f][k+(tt<=T[n-b])][1],tt);
tt=it+L-X[n-b+1]+X[f+1];
mst(dp[b][f+1][k+(tt<=T[f+1])][0],tt);
}
}
cout<<ans;
return 0;
}
컴파일 시 표준 에러 (stderr) 메시지
ho_t3.cpp:19:58: warning: overflow in implicit constant conversion [-Woverflow]
const ll maxn=2e2+10, maxm=1e3+10, lg=21, mod=1e9+7, inf=1e18;
^~~~
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