原题链接

题解

令 \(f(x)\) 代表所有课的发布时间都小于等于x时的不愉快值之和，x越小，AB消耗越大，x越大，C消耗越大，所以感性的想象 \(f(x)\) 是一个下凹函数
然后就可以快乐三分了

code

#define ll unsigned long long
#include<bits/stdc++.h>
using namespace std;

inline void read(ll &x) {
	x = 0;
	ll flag = 1;
	char c = getchar();
    while(c < '0' || c > '9'){
        if(c == '-')flag = -1;
        c = getchar();
    }
	while(c >= '0' && c <= '9') {
		x = (x << 3) + (x << 1) + (c ^ 48);
		c = getchar();
	}
	x *= flag;
}

inline void write(ll x)
{
    if(x < 0){
    	putchar('-');
		x = -x;
	}
    if(x > 9)
		write(x / 10);
    putchar(x % 10 + '0');
}

ll t[100005];
ll b[100005];
ll A,B,C;
ll n,m;

ll f(ll x)
{
    ll sum=0;
    for(ll i=1;i<=n;i++) if(t[i]<x) sum+=C*(x-t[i]);

    if(A>B)
    {
        for(ll i=1;i<=m;i++) if(b[i]>x) sum+=B*(b[i]-x);
    }
    else
    {
        ll extra=0,need=0;

        for(ll i=1;i<=m;i++)
        {
            if(b[i]>x) need+=b[i]-x;//这里对AB的处理太巧妙了，学着点
            else if(b[i]<x) extra+=x-b[i];
        }
        if(need<=extra) sum+=A*need;
        else sum+=A*extra+B*(need-extra);
    }

    return sum;
}

int main()
{
    read(A); read(B); read(C);
    read(n); read(m);
    for(ll i=1;i<=n;i++) read(t[i]);
    for(ll i=1;i<=m;i++) read(b[i]);

    ll l=1,r=100000;
    while(l+10<=r)
    {
        ll midl=l+(r-l)/3;
        ll midr=r-(r-l)/3;

        if(f(midl)>f(midr)) l=midl;
        else r=midr;
    }

    ll ans=1e19;
    for(ll i=l;i<=r;i++) ans=min(ans,f(i));
    write(ans); putchar('\n');
    return 0;
}