抽象出差分约束 然后还有一点就是建立超级源点 优化建图

然后就是比较有趣的拓扑图求差分约束了其实spfa也可

#include<bits/stdc++.h>
using namespace std;
using ll = long long;

const int N = 2e6+10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;

int n,q,m;

int e[N],ne[N],w[N],h[N],idx;
void add(int a,int b,int c){
	e[idx] = b,ne[idx] = h[a],w[idx] = c,h[a] = idx++;
}

bool stop[N];
int din[N];
vector<int>ans;
int ranks[N];


void topsort()
{
	queue<int>q;
	for(int i=1;i<=n;i++)ranks[i] = 1;
	for(int i=1;i<=n+m;++i)if(!din[i]){q.push(i);}
	
	while(q.size()){
		int t = q.front();
		q.pop();
		ans.push_back(t);
		for(int i=h[t];~i;i=ne[i]){
			int j = e[i];
			din[j]--;
			if(!din[j])q.push(j);
		}
	}
	
	
	for(int &ver:ans){
		for(int i=h[ver];~i;i=ne[i]){
			int j = e[i];
			ranks[j] = max(ranks[j],ranks[ver]+w[i]);
		}
	}
	
	int res = 0;
	
	for(int i=1;i<=n;i++)res = max(res,ranks[i]);
	cout<<res;
	
	
	
	
	
}

void solve()
{
	cin>>n>>m;
	memset(h,-1,sizeof h);
	for(int i=1;i<=m;i++){
		int num;cin>>num;
		for(int j=1;j<=n;j++)stop[j] = false;
		int st,ed;st = 0x3f3f3f,ed = 0;
		for(int j=1;j<=num;++j){
			int x;cin>>x;
			stop[x] = true;
			st = min(st,x);
			ed = max(ed,x);
		}
		int ver = n+i;
		
		for(int j=st;j<=ed;++j){
			if(!stop[j]){add(j,ver,0);din[ver]++;}
			else {add(ver,j,1);din[j]++;}
		}
	}
	
	topsort();
	

}

signed main()
{
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int _;
	//cin>>_;
	_ = 1;
	while(_--)solve();
	return 0;
}