题目链接
解题思路
首先，你发现你只能改一个日期，那么我们肯定是改距离最近的旁边的两场考试，此时我们就可以将操作转化为删去一场考试并添加一场新考试的最小的休息时长，容易使用贪心 \(O(n)\) 解决。
总时间复杂度 \(O(n \log_2 n)\)，瓶颈在于排序。
参考代码
#include<bits/stdc++.h>
using namespace std;
//#define map unordered_map
#define forl(i,a,b) for(register long long i=a;i<=b;i++)
#define forr(i,a,b) for(register long long i=a;i>=b;i--)
#define forll(i,a,b,c) for(register long long i=a;i<=b;i+=c)
#define forrr(i,a,b,c) for(register long long i=a;i>=b;i-=c)
#define lc(x) x<<1
#define rc(x) x<<1|1
#define mid ((l+r)>>1)
#define cin(x) scanf("%lld",&x)
#define cout(x) printf("%lld",x)
#define lowbit(x) (x&-x)
#define pb push_back
#define pf push_front
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define endl '\n'
#define QwQ return 0;
#define ll long long
#define ull unsigned long long
#define lcm(x,y) x/__gcd(x,y)*y
#define Sum(x,y) 1ll*(x+y)*(y-x+1)/2
#define aty cout<<"Yes\n";
#define atn cout<<"No\n";
#define cfy cout<<"YES\n";
#define cfn cout<<"NO\n";
#define xxy cout<<"yes\n";
#define xxn cout<<"no\n";
#define printcf(x) x?cout<<"YES\n":cout<<"NO\n";
#define printat(x) x?cout<<"Yes\n":cout<<"No\n";
#define printxx(x) x?cout<<"yes\n":cout<<"no\n";
ll t;
ll n,m;
ll a[200010],b[200010],cz2[200010];
struct node{
ll x,id;
}cz[200010];
bool cmp(node x,node y){
return x.x<y.x;
}
ll f(ll x)// del x
{
ll k=0;
forl(i,1,n)
if(i!=x)
b[++k]=a[i];
ll ans=0;
forl(i,1,n-1)
cz2[i]=b[i]-b[i-1];
forl(i,1,n-1)
ans=max(ans,cz2[i]/2);
ans=max(ans,m-b[n-1]);
sort(cz2+1,cz2+n);
// cout<<ans<<' '<<cz2[1]<<endl;
return min(ans,cz2[1]);
}
void solve()
{
ll ans=0;
cin>>n>>m;
forl(i,1,n)
cin>>a[i],cz[i].x=a[i]-a[i-1],cz[i].id=i;
sort(cz+1,cz+1+n,cmp);
ans=max(ans,f(cz[1].id));
ans=max(ans,f(max(cz[1].id-1,1ll)));
ans=max(ans,f(min(cz[1].id+1,n)));
cout<<ans-1<<endl;
}
int main()
{
IOS;
t=1;
cin>>t;
while(t--)
solve();
QwQ;
}