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  1. #include <bits/stdc++.h>
  2.  
  3. using namespace std;
  4.  
  5. /*
  6. #pragma GCC target ("avx2")
  7. #pragma GCC optimization ("O3")
  8. #pragma GCC optimization ("unroll-loops")
  9. */
  10.  
  11. #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  12. #define rep(i,l,r) for (int i=l;i<=r;i++)
  13. #define fi first
  14. #define se second
  15. #define pb push_back
  16. #define all(x) (x).begin(), (x).end()
  17. #define rall(x) (x).rbegin(), (x).rend()
  18. #define ll long long
  19. #define ull unsigned long long
  20. #define mod 998244353
  21. #define nul(x) memset((x),0,sizeof((x)))
  22.  
  23. mt19937 mrand(random_device{}());
  24. int rnd(int x) { return mrand() % x;}
  25. ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
  26.  
  27. ll n,m,x,k,y;
  28. ll a[200100],b[200100];
  29. vector<ll>v;
  30.  
  31. bool check(){
  32.     ll ind = 1;
  33.     v.pb(0);
  34.     for (int i=1;i<=n;i++){
  35.         if (a[i]==b[ind]){
  36.             ind++;
  37.             v.pb(i);
  38.         }
  39.     }
  40.     v.pb(n+1);
  41.     if (ind==m+1)return true;
  42.     else return false;
  43. }
  44.  
  45. int main(){
  46.     ios;
  47.     memset(a,0,sizeof(a));
  48.  
  49.     cin>>n>>m;
  50.     cin>>x>>k>>y;
  51.     for (int i=1;i<=n;i++){
  52.         cin>>a[i];
  53.     }
  54.     for (int i=1;i<=m;i++){
  55.         cin>>b[i];
  56.     }
  57.  
  58.     if (!check())return cout<<"-1\n",0;
  59.  
  60.     ll sum = 0;
  61.     for (int i=1;i<v.size();i++){
  62.         if (v[i]-v[i-1]==1)continue;
  63.  
  64.         ll mx=0;
  65.         for (int j=v[i-1]+1;j<v[i];j++){
  66.             mx = max(mx,a[j]);
  67.         }
  68.  
  69.         ll ad = 2e18;
  70.         if (max(a[v[i]],a[v[i-1]])>mx){
  71.             ad = min(ad, (ll)y * (ll)(v[i]-v[i-1]-1));
  72.         }
  73.         if ((v[i]-v[i-1]-1) >= k){
  74.             ad = min(ad, (ll)x * (ll)((v[i]-v[i-1]-1)/k) + (ll)y * (ll)((v[i]-v[i-1]-1)%k));
  75.  
  76.             ad = min(ad, x + (ll)y * (ll)((v[i]-v[i-1]-1) - k));
  77.         }
  78.  
  79.         if (ad == 2e18)return cout<<"-1\n",0;
  80.  
  81.         sum += ad;
  82.     }
  83.     cout<<sum;
  84.     return 0;
  85. }
  86.  
Success #stdin #stdout 0s 5136KB
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https://ideone.com/B1cMcc
language:
C++14 (gcc 8.3)
created:
4 years ago
visibility:
public

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