ZOJ-2362 Beloved Sons 贪心 | KM

　　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 ;
 }