传送门
最小割定理我虽然说不太准，但大概就这个意思：对求完最大流后的残图进行tarjin，如果一条边的起点和终点不属于一个强连通分量，则这条边属于最大流。
联系一下，直接A
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define inf 10000000
#include<queue>
using namespace std;
int n,m,S=0,T,e,s,adj[100200],dep[100200];
int zhan[100005],inzhan[100005],dfn[100005],low[100005],tot,head,cnt;
int belong[100005],vis[100005];
struct node
{int u,v,l,next;
} a[400005];
inline int read()
{int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}return x*f;
}
void add(int u,int v,int l)
{a[e].l=l;a[e].v=v;a[e].u=u;a[e].next=adj[u];adj[u]=e++;
}
int bfs()
{memset(dep,0,sizeof(dep));queue<int> q;q.push(S);dep[S]=1;while(!q.empty()){int x=q.front();q.pop();for(int i=adj[x];i!=-1;i=a[i].next){int to=a[i].v;if(!dep[to]&&a[i].l){dep[to]=dep[x]+1;q.push(to);if(to==T)return 1;}}}return 0;
}
int dfs(int x,int len)
{int tmp=len,k;if(x==T)return len;for(int i=adj[x];i!=-1;i=a[i].next){int to=a[i].v;if(dep[to]==dep[x]+1&&a[i].l&&tmp){k=dfs(to,min(a[i].l,tmp));if(!k){dep[to]=0;continue;}a[i].l-=k;a[i^1].l+=k;tmp-=k;}} return len-tmp;
}
void tarjin(int x)
{//vis[x]=1;zhan[++head]=x;inzhan[x]=1;dfn[x]=low[x]=++tot;for(int i=adj[x];i!=-1;i=a[i].next)if(a[i].l&&inzhan[a[i].v]<2){int to=a[i].v;if(!inzhan[to]){tarjin(to);low[x]=min(low[x],low[to]); }else if(inzhan[a[i].v]==1)low[x]=min(low[x],dfn[a[i].v]);}if(dfn[x]==low[x]){cnt++;int tmp;while(1){tmp=zhan[head--];belong[tmp]=cnt;inzhan[tmp]=2;if(tmp==x)break;}}
}
int yjn()
{freopen("sphere.in","r",stdin);freopen("sphere.out","w",stdout);scanf("%d%d",&n,&m);memset(adj,-1,sizeof(adj));int x,y;for(int i=1;i<=m;i++){x=read();y=read();add(x,y,1);add(y,x,0);}T=n+1;for(int i=1;i<=n;i++)if(i<=n/2)add(S,i,1),add(i,S,0);else add(i,T,1),add(T,i,0);while(bfs())dfs(S,inf);for(int i=S;i<=T;i++)if(inzhan[i]<2)tarjin(i);s=0;for(int i=0;i<m;i++)if(a[i*2].l==0&&belong[a[i*2].u]!=belong[a[i*2].v])s++;cout<<s;
}
int qty=yjn();
int main(){;}