思路是每次离开的时候算出到前面和后面保密室的代价，哪边少就往哪边。

实现就是开一个树状数组记录沿过道的学生人数，然后同一行的就直接暴力算。

不知道是思路还是代码有问题。

//A tree without skin will surely die. 
//A man without face is invincible.
#include<bits/stdc++.h>
#define int long long
#define rint register int
using namespace std;
int const N=1e5+10;
int a[N][10],n;
struct Tree_Array{
    int c[N<<2];
    inline int lowbit(int x){return x&-x;}
    inline void update(int x,int v){while (x<=n) c[x]+=v,x+=lowbit(x);}
    inline int query(int x){int ans=0;while (x) ans+=c[x],x-=lowbit(x);return ans;}
}T;
signed main(){
    ios::sync_with_stdio(false);
    cout.tie(0),cout.tie(0);
    int m,A,B;cin>>n>>m>>A>>B;
    for (rint i=1;i<=n;++i)
        for (rint j=1;j<=6;++j)
            a[i][j]=1;
    for (rint i=1;i<=n;++i) T.update(i,2);
    int front=0,behind=0,ans=0;
    for (rint i=1;i<=m;++i){
        string s;cin>>s;
        int h=s[0]-'0',c=s[1]-'A'+1,sumc=0;
        if (c<=3){
            for (rint j=c+1;j<=4;++j)
                sumc+=a[h][j];
        }else{
            for (rint j=3;j<=c-1;++j)
                sumc+=a[h][j];
        }
        if ((sumc+T.query(h-1))*A+front*B<=(sumc+T.query(n)-T.query(h))*A+behind*B){
            ans+=(sumc+T.query(h-1))*A+front*B;++front;
        }else{
            ans+=(sumc+T.query(n)-T.query(h))*A+behind*B;++behind; 
        }
        if ((c==3 || c==4)&& a[h][c]) T.update(h,-1);
        a[h][c]=0;
    }
    cout<<ans<<'\n';
    return 0;
}