我是参考的这篇题解

为啥枚举时,S<=2而不是S<=N，S<=N会爆

下面是我的代码：

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>

#define INF 0x3f3f3f3f
#define N 210
#define M 5010

using namespace std;

int n,m,a[M],b[M],S,T,ans,minn;
int vis[N],inq[N],tot,pre[N];
int edge[M],len[M],nxt[M],head[N];
queue<int> q;

void add(int x,int y,int z)
{
	edge[++tot]=y;
	nxt[tot]=head[x];
	len[tot]=z;
	head[x]=tot;
	edge[++tot]=x;
	nxt[tot]=head[y];
	head[y]=tot;
	len[tot]=0;//双向边 
}

int read()
{
	int v=0,f=1;char c='_';
	while(c<'0'||c>'9')
	{
		if(c=='-')  f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9')
	{
		v=v*10+c-'0';
		c=getchar();
	}
	return v*f;
}

int BFS()
{
	memset(vis,0,sizeof(vis));
	while(q.size())  q.pop();
	q.push(S);inq[S]=INF;
	vis[S]=1;
	while(q.size())
	{
		int x=q.front();
		q.pop();
		for(int i=head[x];i;i=nxt[i])
		{
			int y=edge[i];
			if(len[i]&&!vis[y])
			{
				pre[y]=i;
				inq[y]=min(inq[x],len[i]);//flow=min(len[i],flow);
				if(y==T)  return 1;
				q.push(y);vis[y]=1;
			}
		}
	}
	return 0;
}//不熟练 

void EK()
{
	int x=T;
	while(x!=S)//while(x)
	{
		int y=pre[x];
		len[y]-=inq[T];
		len[y^1]+=inq[T];
		x=edge[y^1];//不会 	x=[x^1];
	}
	minn+=inq[T];
}

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)//cin>>n>>m&&n
	{
		for(int i=1;i<=m;i++)//这儿必须用快读 
		a[i]=read()+1,b[i]=read()+1;//cin>>x>>y;
		ans=INF;
		for(S=1;S<=2;S++)//S<=N
		for(T=1;T<=n;T++)
		if(S!=T)
		{
			memset(head,0,sizeof(head));tot=1;//tot=0;
			for(int i=1;i<=n;i++)
			{
				if(i==S||i==T)  add(i,i+n,INF);
				else  add(i,i+n,1);
			}
			for(int i=1;i<=m;i++)
			{
				add(b[i]+n,a[i],INF);
				add(a[i]+n,b[i],INF);
			}
			minn=0;
			while(BFS())  EK();
			ans=min(ans,minn);
		}
		if(ans==INF||n<=1)  cout<<n<<endl;//???
		else  cout<<ans<<endl;	
	}
	
	return 0;
}