#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <stack>#include <vector>using namespace std;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << (#x) << " = " << (x) << endl#define printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }#define printarr1(a, b) for1(_, 1, b) cout << a[_] << ‘\t‘; cout << endlinline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }inline const int max(const int &a, const int &b) { return a>b?a:b; }inline const int min(const int &a, const int &b) { return a<b?a:b; }const int N=1005, M=1000005;int ihead[N], n, m, cnt, LL[N], FF[N], mp[N][N], tot, s[M<<1], top, vis[N], ok[N], col[N];struct ED { int from, to, next; } e[M<<1];void add(int u, int v) {	e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; e[cnt].from=u;	e[++cnt].next=ihead[v]; ihead[v]=cnt; e[cnt].to=u; e[cnt].from=v;}bool ifind(int u) {	int v;	for(int i=ihead[u]; i; i=e[i].next) if(vis[v=e[i].to]) {		if(col[v]==-1) { col[v]=!col[u]; return ifind(v); }		else if(col[v]==col[u]) return true;	}	return false;}void color(int x) {	int y, u=e[x].from;	CC(vis, 0); CC(col, -1); col[u]=0;	do {		y=s[top--];		vis[e[y].from]=vis[e[y].to]=1;	} while(y!=x);	if(ifind(u)) for1(i, 1, n) if(vis[i]) ok[i]=1;}void tarjan(int u, int fa) {	FF[u]=LL[u]=++tot;	for(int i=ihead[u]; i; i=e[i].next) if(fa!=e[i].to) {		int v=e[i].to;		if(!FF[v]) {			s[++top]=i; //入栈这里要注意。。不要在上边入栈。。			tarjan(v, u);			if(LL[v]>=FF[u]) color(i);			LL[u]=min(LL[u], LL[v]);		}		else if(LL[u]>FF[v]) s[++top]=i, LL[u]=FF[v]; //入栈这里要注意。。	}}int main() {	while(1) {		read(n); read(m); int ans=0;		if(n==0 && m==0) break;		CC(mp, 0); CC(ihead, 0); CC(LL, 0); CC(FF, 0); CC(ok, 0); top=cnt=tot=0;		rep(i, m) {			int u=getint(), v=getint();			mp[u][v]=mp[v][u]=1;		}		for1(i, 1, n) for1(j, i+1, n) if(!mp[i][j]) add(i, j);		for1(i, 1, n) if(!FF[i]) tarjan(i, -1);		for1(i, 1, n) if(!ok[i]) ++ans;		printf("%d\n", ans);	}	return 0;}

• The knights should be seated such that two knights who hate each other should not be neighbors at the table. (Merlin has a list that says who hates whom.) The knights are sitting around a roundtable, thus every knight has exactly two neighbors.
• An odd number of knights should sit around the table. This ensures that if the knights cannot agree on something, then they can settle the issue by voting. (If the number of knights is even, then itcan happen that ``yes" and ``no" have the same number of votes, and the argument goes on.)