uva 1627 Team them up! 二分图 + 连通分量 + 背包

//	uva 1627 Team them up! 二分图 + 连通分量 + 背包
//
//	解题思路:
//		
//		同组之内每个人相互认识,如果a和b不相互认识,则
//	a和b只能分在两个不同的组.所以以不相互认识作为边
//	建图,对于每个联通块判断是否是二分图.如果不是,没有
//	方案,如果有,那么进行背包.
//		d[i][j]表示前i个连通块第一个队伍人数比第二个
//	队伍的人数多j个人是否合法.
//		推出if (d[i][j]) { d[i+1][j+diff[i]] = 1
//						   d[i+1][j-diff[i]] = 1}
//		因为有负数所以加上一个偏移量就可以了.n足矣.
//		
//		初始状态下d[0][0] = 1.现在第二维加上n就ok了
//		然后逆推打印.
//
//		还是很巧妙的,继续加油哟~~~FIGHTING!!!

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <vector>

#define cls(x,a) memset(x,a,sizeof(x))

using namespace std;

const int MAXN = 108;
int n;
int color[MAXN];
int cc;
int g[MAXN][MAXN];
int d[MAXN][MAXN*2];
vector<int> G[MAXN];
vector<int> term[MAXN][2];
bool vis[MAXN];
int diff[MAXN];
void input(){
	scanf("%d",&n);
	cls(g,0);
	cls(color,0);
	cls(vis,0);
	cls(d,0);
	diff[MAXN];
	for (int i = 0 ;i <= n;i ++){
		term[i][0].clear();
		term[i][1].clear();
		G[i].clear();
	}

	for (int i = 1;i <= n;i++){
		int v;
		while(scanf("%d",&v)!=EOF && v){
			g[i][v] = 1;
		}
	}
	if (n == 1){
		puts("No solution");
		return ;
	}
	for (int i=1;i<=n;i++){
		for (int j=1;j<=n;j++){
			if (i != j &&(!g[i][j] || !g[j][i])){
				G[i].push_back(j);
				G[j].push_back(i);
			}
		}
	}

	

}

void p(){
	for (int i = 0 ;i < cc; i++){
		for (int j = 0; j < 2 ; j++){
			for (int k = 0 ; k < term[i][j].size(); k++)
				printf("%d ",term[i][j][k]);
			puts("");
		}
		puts("");
	}
}

bool dfs(int u,int c){
	color[u] = c;
	vis[u] = 1;
	term[cc][c-1].push_back(u);
	for (int i=0;i < G[u].size();i ++){
		int v = G[u][i];
		
		if (v == u) continue;

		if (color[v] && color[u] == color[v]) 	return false;
		if (!color[v] && !dfs(v,3-c))	return false;
	}
	return true;
}

void print(int ans){
	vector<int> term1,term2;
	int t;
	for (int i = cc - 1;i >= 0 ;i --){
		if (d[i][ans-diff[i]+n]){
			t = 0;
			ans -= diff[i];
		}else {
			t = 1;
			ans += diff[i];
		}

		for (int j = 0; j < term[i][t].size(); j++){
			term1.push_back(term[i][t][j]);
		}

		for (int j = 0;j < term[i][t^1].size(); j++)
			term2.push_back(term[i][t^1][j]);
	}

	printf("%d",term1.size());

	for (int j = 0;j < term1.size(); j++)
		printf(" %d",term1[j]);


	puts("");

	printf("%d",term2.size());
	

	for (int j = 0;j < term2.size(); j++)
		printf(" %d",term2[j]);
	
	puts("");

}

void DP(){
	d[0][n] = 1;
	for (int i = 0;i < cc;i ++){
		for (int j=-n ;j <= n;j ++){
			if (d[i][j+n]){
				d[i+1][j+diff[i]+n] = 1;
				d[i+1][j-diff[i]+n] = 1;
			}
		}
	}
//	p();	
	for (int i = 0 ;i <= n;i ++){
		if (d[cc][i+n]){
			print(i);
			return ;
		}
		if (d[cc][-i+n]){
			print(-i);
			return ;
		}
	}


}

void solve(){
	int flag = 0;
	cc = 0;
	for (int i = 1;i <= n;i ++){
		if (!color[i]){
			if (!dfs(i,1)){
				flag = 1;
				break;
			}
			diff[cc] = term[cc][0].size() - term[cc][1].size();
			cc++;
		}
	}
	if (flag){
		puts("No solution");
		return ;
	}
	DP();
}

int main(){
	int t;
	//freopen("1.txt","r",stdin);
	scanf("%d",&t);
	while(t--){
		input();
		solve();
		puts("");
	}
	return 0;
}