uva 10410

這題挺複雜的，給你DFS跟BFS，要你構出一個樹

而且這個樹不一定是二元，可以是多元。

題目也有說如果有很多個答案，印出一種就可以了

利用BFS特性分割，並且由DFS看出每個父節點擁有的子節點

//====================================================================||
//                                                                    ||
//                                                                    ||
//                         Author : GCA                               ||
//                  6AE7EE02212D47DAD26C32C0FE829006                  ||
//====================================================================||
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
using namespace std;
#ifdef ONLINE_JUDGE
#define ll "%lld"
#else
#define ll "%I64d"
#endif
typedef unsigned int uint;
typedef long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\n")
#define I_de(x,n)for(int i=0;i<n;i++)printf("%d ",x[i]);Pln()
#define De(x)printf(#x"%d\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\
    printf("%d ",dp[htx][hty]);}Pln();}
#define M 1005
#define PII pair<int,int>
#define PB push_back
#define oo INT_MAX
#define Set_oo 0x3f
#define Is_debug true
#define debug(...) if(Is_debug)printf("DEBUG: "),printf(__VA_ARGS__)
#define FOR(it,c) for(typeid((c).begin()) it=(c).begin();it!=(c).end();it++)
#define eps 1e-6
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
    int tmp=min(x,y);
    return min(tmp,z);
}
int max3(int x,int y,int z){
    int tmp=max(x,y);
    return max(tmp,z);
}
int n;
int x[M];
int y[M];
vector<int> ans[M];
vector<int> pos[M];
vector<int> have[M];
void bt(int h,int s){
//    debug("%d\n",h);
//    for(int i=1;i<=n;i++){
//        printf("%d :",i);
//        for(int j=0;j<have[i].size();j++){
//            printf("%d ",have[i][j]);
//        }Pln();
//    }Pln();
    int record=-1;
    int rej=-1;
    bool f=true;
    for(int i=s;i<n;i++){
        bool ha=false;
        int j;
        for(j=0;j<have[h].size();j++){
            if(have[h][j]==x[i]){
                ha=true;
                break;
            }
        }
        if(ha&&x[i]>record&&j>rej){
            f=false;
            record=x[i];
            rej=j;
            ans[h].PB(x[i]);
            pos[h].PB(j);
        }
        else if(!f)break;
    }
    for(int i=0;i<pos[h].size();i++){
        int tmp=(i==pos[h].size()-1)?have[h].size():pos[h][i+1];
        for(int j=pos[h][i]+1;j<tmp;j++){
            have[ans[h][i]].PB(have[h][j]);
        }
        bt(ans[h][i],s+pos[h].size());
    }

    return ;
}
void solve(){
    for(int i=1;i<n;i++){
        have[x[0]].PB(y[i]);
    }
    bt(x[0],1);
    for(int i=1;i<=n;i++){
        printf("%d:",i);
        for(int j=0;j<ans[i].size();j++)printf(" %d",ans[i][j]);
        Pln();
    }
}
void clear(){
    for(int i=0;i<=n;i++){
        have[i].clear();
        ans[i].clear();
        pos[i].clear();
    }
}
int main() {
    ios_base::sync_with_stdio(0);
    while(~scanf("%d",&n)){
        for(int i=0;i<n;i++)scanf("%d",&x[i]);
        for(int i=0;i<n;i++)scanf("%d",&y[i]);
        solve();
        clear();
    }
















}