Codeforces 1091E New Year and the Acquaintance Estimation [图论]

洛谷

Codeforces

---

思路

有一个定理：Erdős–Gallai定理。

然后观察样例，可以猜到答案必定是奇偶性相同的一段区间，那么二分左右端点即可。

定理和这个猜测暂时都懒得学/证，留坑。

---

#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
	using namespace std;
	#define pii pair<int,int>
	#define fir first
	#define sec second
	#define MP make_pair
	#define rep(i,x,y) for (int i=(x);i<=(y);i++)
	#define drep(i,x,y) for (int i=(x);i>=(y);i--)
	#define go(x) for (int i=head[x];i;i=edge[i].nxt)
	#define templ template<typename T>
	#define sz 505005
	typedef long long ll;
	typedef double db;
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
	templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
	templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
	templ inline void read(T& t)
	{
		t=0;char f=0,ch=getchar();double d=0.1;
		while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
		while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
		if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
		t=(f?-t:t);
	}
	template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
	char sr[1<<21],z[20];int C=-1,Z=0;
    inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
    inline void print(register int x)
    {
    	if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
    	while(z[++Z]=x%10+48,x/=10);
    	while(sr[++C]=z[Z],--Z);sr[++C]='\n';
    }
	void file()
	{
		#ifndef ONLINE_JUDGE
		freopen("a.in","r",stdin);
		#endif
	}
	inline void chktime()
	{
		#ifndef ONLINE_JUDGE
		cout<<(clock()-t)/1000.0<<'\n';
		#endif
	}
	#ifdef mod
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
	ll inv(ll x){return ksm(x,mod-2);}
	#else
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
	#endif
//	inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;

int n;
int aa[sz],a[sz];
ll s[sz];

int check(int w) // return value : 0:ok;1:big;-1:small
{
	int n=::n+1,t=1;
	rep(i,1,n-1) a[i]=aa[i];a[n]=w;
	drep(i,n,2)
		if (a[i-1]<=a[i]) swap(a[i-1],a[i]);
		else {t=i;break;}
	drep(i,n,1) s[i]=s[i+1]+a[i];
	int p=n+1;
	rep(k,1,n)
	{
		while (p>1&&a[p-1]<=k) --p;
		ll S;
		if (p<=k+1) S=1ll*k*(k-1)+s[k+1];
		else S=1ll*k*(k-1)+s[p]+1ll*k*(p-1-(k+1)+1);
		if (s[1]-s[k+1]<=S) continue;
		return p>t?1:-1;
	}
	return 0;
}

int main()
{
	file();
	int s=0;
	read(n);
	rep(i,1,n) read(aa[i]),s+=(aa[i]&1);
	sort(aa+1,aa+n+1);reverse(aa+1,aa+n+1);
	s&=1;
	int l,r,L=-1,R=-1;
	l=0,r=n/2;
	while (l<=r)
	{
		int mid=(l+r)>>1;
		if (check(mid*2+s)>=0) L=mid*2+s,r=mid-1;
		else l=mid+1;
	}
	l=0,r=n/2;
	while (l<=r)
	{
		int mid=(l+r)>>1;
		if (check(mid*2+s)<=0) R=mid*2+s,l=mid+1;
		else r=mid-1;
	}
	if (L==-1||R==-1||L>R) puts("-1");
	else for (int i=L;i<=R;i+=2) printf("%d ",i);
	return 0;
}