题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2362
裸的匹配问题,直接KM,就算是O(n^4)的KM也不会超。当然注意到题目中左边的点到右点所连的边的权值是一样的,所以完全可以贪心拍个序,然后找增广路。。。
//STATUS:C++_AC_250MS_848KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
//typedef __int64 LL;
//typedef unsigned __int64 ULL;
//const
const int N=;
const int INF=0x3f3f3f3f;
const int MOD=,STA=;
//const LL LNF=1LL<<60;
const double EPS=1e-;
const double OO=1e15;
const int dx[]={-,,,};
const int dy[]={,,,-};
const int day[]={,,,,,,,,,,,,};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End struct Node{
int val,id;
bool operator < (const Node& a)const{
return val>a.val;
}
}nod[N];
int ca; int w[N][N],y[N],vis[N];
int n,m; int dfs(int u)
{
int v;
for(v=;v<=n;v++){
if(w[u][v] && !vis[v]){
vis[v]=;
if(y[v]==- || dfs(y[v])){
y[v]=u;
return ;
}
}
}
return ;
} int main()
{
// freopen("in.txt","r",stdin);
int i,j,tot,a;
int x[N];
scanf("%d",&ca);
while(ca--)
{
scanf("%d",&n);
for(i=;i<=n;i++){
scanf("%d",&nod[i].val);
nod[i].val*=nod[i].val;
nod[i].id=i;
}
mem(w,);
for(i=;i<=n;i++){
scanf("%d",&tot);
while(tot--){
scanf("%d",&a);
w[i][a]=nod[i].val;
}
}
sort(nod+,nod+n+); mem(x,);
mem(y,-);
for(i=;i<=n;i++){
mem(vis,);
dfs(nod[i].id);
}
for(i=;i<=n;i++)
if(y[i]!=-)x[y[i]]=i; printf("%d",x[]);
for(i=;i<=n;i++)
printf(" %d",x[i]);
putchar('\n');
}
return ;
}