晕，m=0是要输出1（弄的我还找管理员要数据，但明显题意是叫我们输出0呀）
最大密度子图，把边转换成点，然后二分答案，跑最大权闭合子图判定是否可行。
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#define N 1110
#define oo 0x3f3f3f3f
using namespace std; struct Edge {
int u, v, f;
Edge( int u, int v, int f ):u(u),v(v),f(f){}
};
int gcd( int a, int b ) {
return b?gcd(b,a%b):a;
}
struct Pair {
int a, b;
Pair( int aa, int bb ) {
int cd = gcd(aa,bb);
a=aa/cd;
b=bb/cd;
}
bool operator<( const Pair &o ) const {
return a*o.b < o.a*b;
}
bool operator==( const Pair &o ) const {
return a*o.b==o.a*b;
}
};
int n, m;
vector<Edge> edge;
vector<int> g[N];
vector<Pair> prs;
int dep[N], cur[N], qu[N], bg, ed;
int uu[N], vv[N], idx[N], src, dst, idc;
bool vis[N]; void init() {
for( int i=; i<=dst; i++ )
g[i].clear();
edge.clear();
}
void adde( int u, int v, int f ) {
g[u].push_back( edge.size() );
edge.push_back( Edge(u,v,f) );
g[v].push_back( edge.size() );
edge.push_back( Edge(v,u,) );
}
bool bfs() {
memset( dep, , sizeof(dep) );
qu[bg=ed=] = src;
dep[src] = ;
while( bg<=ed ) {
int u=qu[bg++];
for( int t=; t<g[u].size(); t++ ) {
Edge &e=edge[g[u][t]];
if( e.f && !dep[e.v] ) {
dep[e.v]=dep[e.u]+;
qu[++ed] = e.v;
}
}
}
return dep[dst];
}
int dfs( int u, int a ) {
if( u==dst || a== ) return a;
int remain=a, past=, na;
for( int &t=cur[u]; t<g[u].size(); t++ ) {
Edge &e=edge[g[u][t]];
Edge &ve=edge[g[u][t]^];
if( e.f && dep[e.v]==dep[e.u]+ && (na=dfs(e.v,min(e.f,remain))) ) {
remain -= na;
past += na;
e.f -= na;
ve.f += na;
if( !remain ) break;
}
}
return past;
}
int maxflow() {
int flow = ;
while( bfs() ) {
memset( cur, , sizeof(cur) );
flow += dfs(src,oo);
}
return flow;
}
void makeid() {
src = n+m+;
dst = src+;
idc = n;
for( int i=; i<=m; i++ ) idx[i] = ++idc;
}
void rebuild( int a, int b ) {
init();
for( int i=; i<=m; i++ ) {
adde( src, idx[i], b );
adde( idx[i], uu[i], oo );
adde( idx[i], vv[i], oo );
}
for( int i=; i<=n; i++ )
adde( i, dst, a );
}
bool ok( int a, int b ) {
rebuild(a,b);
return m*b-maxflow() > ;
}
int calc() {
int cnt = ;
qu[bg=ed=] = dst;
vis[dst] = true;
while( bg<=ed ) {
int u=qu[bg++];
for( int t=; t<g[u].size(); t++ ) {
Edge &e = edge[g[u][t]^];
if( e.f && !vis[e.u] ) {
vis[e.u] = true;
cnt += <=e.u&&e.u<=n;
qu[++ed] = e.u;
}
}
}
return n-cnt;
}
int binary() {
for( int a=; a<=m; a++ )
for( int b=; b<=n; b++ )
prs.push_back( Pair(a,b) );
sort( prs.begin(), prs.end() );
prs.erase( unique(prs.begin(),prs.end()), prs.end() );
int lf=, rg=prs.size()-;
while( lf<rg ) {
int mid = (lf+rg)>>;
if( ok(prs[mid].a,prs[mid].b) )
lf=mid+;
else
rg=mid;
}
ok(prs[lf].a,prs[lf].b);
return calc();
}
int main() {
scanf( "%d%d", &n, &m );
if( m== ) {
printf( "1\n" );
return ;
}
for( int i=; i<=m; i++ )
scanf( "%d%d", uu+i, vv+i );
makeid();
printf( "%d\n", binary() );
}