xg

題意

　　有m種花，每種花數量無限。每個花有兩個幸福值a，b，一種花的第一朵花是a，其餘全是b。問買n朵花，最高幸福值是多少。

思路

　　對於每種花。第i種花，n朵花全買這種，然後二分查詢a，找到比bi大的a，然後計算數量cnt（cnt的大小需要仔細斟酌）。n中買i的cnt個不買i了，改買比bi大的a。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include 
 
<iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '\n'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define
 
 bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define
 
 sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 2e5+10;
ll n,m,x,k,y;
int p[maxn],a[maxn],b[maxn];
int tail;
queue<pair<int,int> >que;
void Init()
{
    cin>>n>>m;
    cin>>x>>k>>y;
    rep(i,1,n) cin>>a[i];
    rep(i,1,m) cin>>b[i];    
    int flag = 1;
    int pos = 1;
    tail = 0;
    rep(i,1,n){
        if(a[i] == b[pos]){
            p[++tail] = i;
            pos++;
        }
    }
    if(tail != m){
        cout<<-1<<endl;
        exit(0);
    } 
    int maxx = a[p[1]];
    int cnt = 0;
    rep(i,1,p[1]-1){
        if(a[i] > maxx){
            cnt++;
        }
    }
    que.push(mp(cnt,p[1]-1));
    rep(i,2,tail){
        int l = p[i-1];
        int r = p[i];
        maxx = max(a[l],a[r]);
        cnt = 0;
        rep(j,l+1,r-1){
            if(a[j] > maxx) cnt++;
        }
        que.push(mp(cnt,r-l-1));
    }
    maxx = a[p[tail]];
    cnt = 0;
    rep(i,p[tail]+1,n){
        if(a[i] > maxx) cnt++;
    }
    que.push(mp(cnt,n-p[tail]));

}
ll ans = 0;
void solve()
{
    while(que.size()){
        int len = que.front().se;
        int flag = que.front().fi;
        que.pop();
         if(len < k){
            if(flag) {
                cout<<-1<<endl;
                exit(0);
            }
            else {
                ans += len*y;
            }
        }
        else {
            int tmp1 = len/k;
            int tmp2 = len%k;
            if(flag){
                ll sum1 = x + (len-k)*y;
                ll sum2 = tmp1*x + tmp2*y;
                ans += min(sum1,sum2);
            }else{
                ll sum1 = len*y;
                ll sum2 = tmp1*x + tmp2*y;
                ans += min(sum1,sum2);
            }
        }
    }
}

int main()
{
    //freopen("input.txt", "r", stdin);
    Init();
    solve();
    cout<<ans<<endl;

}