题解 给国与时光机

传送门

不想写这个题解了
全是人类智慧
做法见题解

一个有趣的事情是关于无解的证明，十分巧妙：

为啥逆序对是奇数就一定不是一个大环呢？
考虑全是自环的情况逆序对为 0
每次可以以逆序对数 +1 为代价合并两个环
所以只有一个大环一定合并了偶数次
（大概）

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define fir first
#define sec second
#define pb push_back
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, m;
int a[200010];

namespace task{
	int top;
	bool vis[200010];
	pair<int, int> sta[200010];
	vector<int> N, A, L, R, tem;
	void solve() {
		for (int i=1; i<=m; ++i) vis[a[i]]=1;
		for (int i=1; i<=2*n; ++i) {
			if (!vis[i-1]&&!vis[i]) N.pb(i);
			else if (vis[i-1]&&!vis[i]) L.pb(i);
			else if (!vis[i-1]&&vis[i]) R.pb(i);
			else A.pb(i);
		}
		// cout<<"siz: "<<N.size()<<' '<<L.size()<<' '<<R.size()<<' '<<A.size()<<endl;
		if ((N.size()|A.size())&1) {puts("No"); exit(0);}
		for (int i=0; i<N.size(); i+=2) sta[++top]={N[i], N[i+1]};
		if (A.size()%4==0) {
			puts("Yes");
			assert(L.size()==R.size());
			for (int i=0; i<L.size(); ++i) sta[++top]={L[i], R[i]};
			for (int i=0; i<A.size(); i+=4) {
				sta[++top]={A[i], A[i+2]};
				sta[++top]={A[i+1], A[i+3]};
			}
			for (int i=1; i<=top; ++i) printf("%d %d\n", sta[i].fir, sta[i].sec);
		}
		else {
			if (L.size()<2) {puts("No"); exit(0);}
			int fir_a=0, sec_a=0, pos=0;
			for (auto it:A) if (it>R[0]) {fir_a=it; break;}
			if (!fir_a) {puts("No"); exit(0);}
			for (int i=0; i<L.size(); ++i) if (L[i]>fir_a) {pos=i; break;}
			if (!pos) {puts("No"); exit(0);}
			for (auto it:A) if (it>R[pos]) {sec_a=it; break;}
			if (!sec_a) {puts("No"); exit(0);}
			// cout<<fir_a<<' '<<sec_a<<' '<<pos<<endl;
			sta[++top]={L[0], R[pos]};
			sta[++top]={R[0], L[pos]};
			sta[++top]={fir_a, sec_a};
			for (int i=1; i<L.size(); ++i) if (i!=pos)
				sta[++top]={L[i], R[i]};
			for (auto it:A) if (it!=fir_a&&it!=sec_a) tem.pb(it);
			for (int i=0; i<tem.size(); i+=4) {
				sta[++top]={tem[i], tem[i+2]};
				sta[++top]={tem[i+1], tem[i+3]};
			}
			puts("Yes");
			for (int i=1; i<=top; ++i) printf("%d %d\n", sta[i].fir, sta[i].sec);
		}
	}
}

signed main()
{
	freopen("shuttle.in", "r", stdin);
	freopen("shuttle.out", "w", stdout);

	n=read(); m=read();
	for (int i=1; i<=m; ++i) a[i]=read();
	task::solve();

	return 0;
}