当你的二分图统计的是差的个数，考虑答案是否统计错误，建议解个方程

#include<bits/stdc++.h>
#define int long long
using namespace std;
namespace wxh666{
	#define getchar getchar_unlocked
	#define putchar putchar_unlocked
	#define sqrt(a) __builtin_sqrt(a)
	#define f(a,b,c) for(int a=b;a<=c;a++)
	#define ff(a,b,c) for(int a=b;a>=c;a--)
	#define max(x,y) (x>y?x:y)
	#define min(x,y) (x<y?x:y)
    template<typename T>
    inline void read(T &ans)
    {
		char ch=getchar();int f=1;ans=0;
		for(;!isdigit(ch);ch=getchar()) ch=='-'?f=-1:f=1;
		for(;isdigit(ch);ch=getchar()) ans=(ans<<3)+(ans<<1)+(ch&15);
		ans*=f;
        return;
    }
    template<typename T,typename ...Args>
    inline void read(T &tmp,Args &...tmps){read(tmp);read(tmps...);}
	inline int read()
	{
		char ch=getchar();int f=1,ans=0;
		for(;!isdigit(ch);ch=getchar()) ch=='-'?f=-1:f=1;
		for(;isdigit(ch);ch=getchar()) ans=(ans<<3)+(ans<<1)+(ch&15);
		return f*ans;
	}
	#define in read()
    template<typename T>
	inline void p(T x,bool g=0)
	{
		ios::sync_with_stdio(false);
		cout<<x;
		ios::sync_with_stdio(true);
		return;
	}
    template<typename T>
	inline void pk(T x,bool g=0)
	{
		ios::sync_with_stdio(false);
		cout<<x<<" ";
		ios::sync_with_stdio(true);
		return;
	}
    template<typename T,typename ...Args>
	inline void p(T x,Args ...xs){p(x);p(xs...);}
    template<typename T,typename ...Args>
	inline void pk(T x,Args ...xs){pk(x);pk(xs...);}
	#define cs(dt,n,m) \
    for(int i=1;i<=n;i++)\
    {\
        for(int j=1;j<=n;j++)\
            printf("%d ",dt[i][j]);\
        cout<<endl;\
    }
};using namespace wxh666;
namespace lsq {
    typedef int lsqxx;
    struct lq {
        struct lqbz {
            lsqxx v,w,nxt;
        }e[8000005];
        lsqxx h[2005],cnt;
        inline void add(lsqxx u,lsqxx v,lsqxx w=1) {
            e[++cnt].v=v;
            e[cnt].w=w;
            e[cnt].nxt=h[u];
            h[u]=cnt;
        }
    void erase() {cnt=0;memset(h,0,sizeof(h));return;}
    #define F(z,u) for(int j=z.h[u],v=z.e[j].v,w=z.e[j].w;j;j=z.e[j].nxt,v=z.e[j].v,w=z.e[j].w)
    }q;
};
using namespace lsq;
int n;
int dt[2005][2005],x;
int color[2005];
int a[2005],acnt;
void dfs(int u,int val)
{
    a[acnt]+=(val<<1)-1;
    color[u]=val;
    F(q,u)
    {
        if(color[v]==val)
        {
            puts("No solution");
            exit(0);
        }
        if(color[v]!=-1) continue;
        dfs(v,val^1);
    }
}
int dp[2005][4005];
#define dp(i,j) dp[i][j+2000]
signed main()
{
	cin>>n;
    f(i,1,n)
    {
        color[i]=-1;
        while(true)
        {
            x=read();
            if(!x) break;
            dt[i][x]=1;
        }
    }
    f(i,1,n) f(j,i+1,n) if((!dt[i][j])||(!dt[j][i])) q.add(i,j),q.add(j,i);
    f(i,1,n) if(color[i]==-1) acnt++,dfs(i,1),a[acnt]=abs(a[acnt]);
    dp(0,0)=1;
    f(i,1,acnt)
        f(j,-n,n)
            dp(i,j)=dp(i-1,j-a[i])|dp(i-1,j+a[i]);
    f(i,0,n)
    {
        if(dp(acnt,i))
        {
            pk((n-i)/2,(n+i)/2,'\n');
            return 0;
        }
    }
	return 0;
}
/*
a+b=n
b-a=i

b=(n+i)/2
a=(n-i)/2


dp[i][j] 前i个块的集合点数差为j是否可以达到
*/