F. Tree Restoration-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_39132605/article/details/101925149

本文介绍了一种基于图的拓扑排序算法实现方法，并通过一个具体的编程实例展示了如何使用队列来完成这一过程。该算法能够判断一个有向无环图是否可以进行拓扑排序，并输出相应的排序结果。

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http://codeforces.com/gym/101755/problem/F

#include <iostream>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <set>
#include <map>
#include <stack>
//#include <tr1/unordered_map>
//#include <unordered_map>
#include <cmath>
//#include<bits/stdc++.h>
using namespace std;

#define sfi(i) scanf("%d",&i)
//#define sfl(i) scanf("%I64d",&i)
#define sfs(i) scanf("%s",(i))
#define pri(i) printf("%d\n",i)
#define prl(i) printf("%I64d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define ull unsigned long long
#define uint unsigned int
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define inf 8e19
#define eps 1e-10
#define PI acos(-1.0)
#define lowbit(x) ((x)&(-x))
#define fl() printf("flag\n")
#define MOD(x) ((x%mod)+mod)%mod
#define endl '\n'
#define pb push_back
#define lson (rt<<1)
#define rson (rt<<1|1)
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)

template<typename T>inline void read(T &x)
{
    x=0;
    static int p;p=1;
    static char c;c=getchar();
    while(!isdigit(c)){if(c=='-')p=-1;c=getchar();}
    while(isdigit(c)) {x=(x<<1)+(x<<3)+(c-48);c=getchar();}
    x*=p;
}

//-----------------------------------------------
const int maxn=1e3+9;
const int mod=1e9+7;

int du[maxn];
vector<int >g[maxn];
int fa[maxn];
int main()
{
    //FAST_IO;

    //freopen("input.txt","r",stdin);

    int n;
    while(~sfi(n))
    {
        for(int i=1;i<=n;i++) du[i]=0,fa[i]=-1,g[i].clear();

        for(int i=1;i<=n;i++)
        {
            int c;
            sfi(c);
            while(c--)
            {
                int v;
                sfi(v);
                g[i].pb(v);
                du[v]++;
            }
        }

        queue<int>q;
        for(int i=1;i<=n;i++)
        {
            if(du[i]==0) q.push(i),fa[i]=i;
        }

        bool f=1;
        if(q.size()!=1)
        {
            puts("NO");
            continue;
        }

        int num=0;
        while(!q.empty())
        {
            int u=q.front();
            q.pop();

            for(int i=0;i<g[u].size();i++)
            {
                int v=g[u][i];
                if(fa[v]!=fa[u]&&fa[v]!=-1)
                {
                    f=0;
                }
                fa[v]=u;
                du[v]--;
                if(du[v]==0)
                {
                    q.push(v);
                    num++;
                }
            }
        }

        if(num!=n-1) f=0;

        if(!f) puts("NO");
        else
        {
            puts("YES");

            for(int i=1;i<=n;i++)
            {
                if(fa[i]!=-1&&fa[i]!=i)
                {
                    printf("%d %d\n",fa[i],i);
                }
            }

        }

    }


    return 0;
}