Apartments

合集 - 贪心(11)

1.P2887 [USACO07NOV] Sunscreen G2024-03-282.F. 0, 1, 2, Tree!2024-03-293.P2949 [USACO09OPEN] Work Scheduling G2024-05-204.C. Given Length and Sum of Digits...2024-06-065.C. Lexicographically Largest2024-06-176.B. Neutral Tonality2024-06-177.F. Lisa and the Martians2024-06-178.E. 我要打 k 个2024-06-189.D. Armchairs2024-06-1810.D. Andrey and Escape from Capygrad2024-07-13

11.Apartments2024-07-30

收起

原题链接

题解

考虑从左到右第一个在游客范围里的公寓，显然把这个公寓分给范围里最左边的游客最优，因为右边的游客更高概率遇到公寓

code

#include<bits/stdc++.h>
#define ll long long
#define lowbit(x) ((x)&(-x))
using namespace std;
const ll inf=1e18;
const ll mod=1e9+7;

ll a[2000005],b[200005];

void solve()
{
    ll n,m,k;
    cin>>n>>m>>k;

    for(ll i=1;i<=n;i++)
    {
        cin>>a[i];
    }

    for(ll i=1;i<=m;i++)
    {
        cin>>b[i];
    }

    sort(a+1,a+1+n);
    sort(b+1,b+1+m);

    ll it1=1;
    ll ans=0;
    for(ll it2=1;it2<=m;it2++)
    {
        while(it1<=n&&a[it1]<b[it2]-k) it1++;

        if(it1>n) break;
        if(a[it1]>b[it2]+k) continue;

        ans++;
        it1++;
    }
    cout<<ans;
}
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int TT=1;
    //cin>>TT;
    while(TT--) solve();
    return 0;
}