codeforces（E. Restorer Distance）三分

最新推荐文章于 2022-07-13 00:03:04 发布

路西菲尔

最新推荐文章于 2022-07-13 00:03:04 发布

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分类专栏：题解

本文链接：https://blog.youkuaiyun.com/qq_43781431/article/details/106176448

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凹函数，三分高度即可

代码：

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

ll e[200005],sum[200005];
ll ans=9223372036854775807,n,a,r,m;
ll f(ll x)
{
    ll pos=lower_bound(e+1,e+1+n,x)-e;
    ll high=sum[n]-sum[pos-1]-x*(n-pos+1);
    ll low=x*(pos-1)-sum[pos-1];
    if(a+r<=m)
    {return low*a+high*r;}
    else
    {
        if(high>=low)
        {return low*m+(high-low)*r;}
        else
        {return high*m+(low-high)*a;}
    }
}
int main()
{
    scanf("%lld %lld %lld %lld",&n,&a,&r,&m);
    for(int i=1;i<=n;i++)
    {scanf("%lld",&e[i]);}
    if(n==1){printf("0\n");return 0;}
    sort(e+1,e+1+n);
    for(int i=1;i<=n;i++)
    {
        sum[i]=sum[i-1]+e[i];
    }
    ll ans1=f(0),ans2=f(1000000000);
    ans=min(ans,ans1);
    ans=min(ans,ans2);
    ll l=0,r=1000000000,m1,m2;
    for(int i=1;i<=1000;i++)
    {
        ll m1=(l+r)/2;
        ll m2=(m1+r)/2;
        ans1=f(m1);
        ans2=f(m2);
        ans=min(ans1,ans);
        ans=min(ans2,ans);
        if(m1==l&&r==m2)
        {break;}
        if(ans1>ans2)
        {l=m1;}
        else
        {r=m2;}
    }
    printf("%lld\n",ans);
    return 0;
}