Given a dense graph can be divided into no more than two groups, described in terms of its complement. How many seek to meet later point edge built between them to increase the maximum group size will be. \ (n \ leq 10 ^ 4 , m \ leq 1.5 \ times 10 ^ 5 \)
Solution
Maximum picture group is the complement graph maximum independent set, the question is intended to complement FIG bipartite graph, which thus omitted for the sake of conversion so that a maximum independent set of edges may bipartite graph reduction
Considering the maximum independent set number = the maximum number of matches, then for the sake of conversion in which a certain side of a maximum matching in
Theorem of an edge of the bipartite graph constant if and only if this edge matching the maximum full flow, and the remaining amount of the network two vertices of this edge not in the same SCC in
So we ran out of a maximum matching, and residual network and run Tarjan, enumerate each match can be watched and SCC are the same
Note that with Hungary will be T, so run maximum flow
Since the input data points which we do not know which part in, so the first bipartite graph coloring it
I found a strong link board is fake ......
#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
namespace scc {
vector <int> g[N],scc[N];
int ind,f[N],siz[N],dfn[N],low[N],vis[N],s[N],bel[N],top,tot,n,m,d[N];
char ch[N];
void make(int p,int q) {
d[q]++;
g[p].push_back(q);
}
void dfs(int p) {
s[++top]=p;
dfn[p]=low[p]=++ind;
for(int i=0;i<g[p].size();i++) {
int q=g[p][i];
if(!dfn[q]) dfs(q), low[p]=min(low[p],low[q]);
else if(!bel[q]) low[p]=min(low[p],dfn[q]);
}
if(dfn[p]==low[p]) {
++tot;
for(int i=0;i!=p;) {
i=s[top--];
bel[i]=tot;
scc[tot].push_back(i);
}
}
}
void solve(int _n) {
n=_n;
for(int i=1;i<=n;i++) if(!dfn[i]) dfs(i);
}
}
namespace flow {
const int maxn = 500005;
const int inf = 1e+9;
int p1[maxn],p2[maxn],p3[maxn];
int dis[maxn], ans, cnt = 1, s, t, pre[maxn * 10], nxt[maxn * 10], h[maxn], v[maxn * 10];
queue<int> q;
void make(int x, int y, int z) {
pre[++cnt] = y, nxt[cnt] = h[x], h[x] = cnt, v[cnt] = z;
p1[cnt]=x; p2[cnt]=y; p3[cnt]=z;
pre[++cnt] = x, nxt[cnt] = h[y], h[y] = cnt;
p1[cnt]=y; p2[cnt]=x; p3[cnt]=z;
}
bool bfs() {
memset(dis, 0, sizeof dis);
q.push(s), dis[s] = 1;
while (!q.empty()) {
int x = q.front();
q.pop();
for (int i = h[x]; i; i = nxt[i])
if (!dis[pre[i]] && v[i])
dis[pre[i]] = dis[x] + 1, q.push(pre[i]);
}
return dis[t];
}
int dfs(int x, int flow) {
if (x == t || !flow)
return flow;
int f = flow;
for (int i = h[x]; i; i = nxt[i])
if (v[i] && dis[pre[i]] > dis[x]) {
int y = dfs(pre[i], min(v[i], f));
f -= y, v[i] -= y, v[i ^ 1] += y;
if (!f)
return flow;
}
if (f == flow)
dis[x] = -1;
return flow - f;
}
int solve(int _s,int _t) {
s=_s;
t=_t;
ans = 0;
for (; bfs(); ans += dfs(s, inf));
return ans;
}
}
int n, m, s, t, t1, t2, t3, x[N],y[N];
int id[N];
struct pii {
int x,y;
bool operator < (const pii &b) {
if(x==b.x) return y<b.y;
else return x<b.x;
}
};
namespace color {
vector <int> g[N];
int c[N];
void make(int p,int q) {
g[p].push_back(q);
g[q].push_back(p);
}
void dfs(int p) {
for(int q:g[p]) if(c[q]==0) {
c[q]=3-c[p];
dfs(q);
}
}
void solve() {
for(int i=1;i<=n;i++) {
if(c[i]==0) {
c[i]=1;
dfs(i);
}
}
}
}
int p1[N],p2[N];
int main() {
ios::sync_with_stdio(false);
cin>>n>>m;
for(int i=1;i<=m;i++) {
cin>>t1>>t2;
p1[i]=t1;
p2[i]=t2;
color::make(t1,t2);
}
color::solve();
for(int i=1;i<=m;i++) {
t1=p1[i];
t2=p2[i];
if(color::c[p1[i]]==2) swap(t1,t2);
id[i]=flow::cnt+1;
x[i]=t1;
y[i]=t2;
flow::make(t1,t2,1);
}
for(int i=1;i<=n;i++) {
if(color::c[i]==1) flow::make(n+1,i,1);
else flow::make(i,n+2,1);
}
flow::solve(n+1,n+2);
for(int i=1;i<=flow::cnt;i++) {
if(flow::v[i]!=0) scc::make(flow::p1[i],flow::p2[i]);
}
scc::solve(n+2);
vector <pii> vec;
for(int i=1;i<=m;i++) {
if(flow::v[id[i]]==0) if(scc::bel[x[i]]!=scc::bel[y[i]])
vec.push_back({min(x[i],y[i]),max(x[i],y[i])});
}
cout<<vec.size()<<endl;
sort(vec.begin(),vec.end());
for(int i=0;i<vec.size();i++) cout<<vec[i].x<<" "<<vec[i].y<<endl;
}