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=410;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
//const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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=1;v<=n;v++){
        if(w[u][v] && !vis[v]){
            vis[v]=1;
            if(y[v]==-1 || dfs(y[v])){
                y[v]=u;
                return 1;
            }
        }
    }
    return 0;
}

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=1;i<=n;i++){
            scanf("%d",&nod[i].val);
            nod[i].val*=nod[i].val;
            nod[i].id=i;
        }
        mem(w,0);
        for(i=1;i<=n;i++){
            scanf("%d",&tot);
            while(tot--){
                scanf("%d",&a);
                w[i][a]=nod[i].val;
            }
        }
        sort(nod+1,nod+n+1);

        mem(x,0);
        mem(y,-1);
        for(i=1;i<=n;i++){
            mem(vis,0);
            dfs(nod[i].id);
        }
        for(i=1;i<=n;i++)
            if(y[i]!=-1)x[y[i]]=i;

        printf("%d",x[1]);
        for(i=2;i<=n;i++)
            printf(" %d",x[i]);
        putchar('
');
    }
    return 0;
}