Codeforces gym101755F Tree Restoration(拓扑排序)

题意：

一棵树，给出每个点的后代们，问你这棵树是否存在，存在就给出这棵树

n<=1000

思路：

对祖先->后代建立有向图，跑拓扑排序。跑的时候不断更新父亲并判断答案的存在性，同时注意一种情况：一个点他儿子是他的后代，同样也得是他父亲的后代，这样传递下去就一定是所有祖宗的后代。

代码：

代码参考网上的

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x))

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 1e6+;
const int maxm = 1e6+;
//const int inf = 0x3f3f3f3f;
const int inf = 1e9+;
const db pi = acos(-1.0);

int n;
vector<int>v[maxn];
int f[maxn];
int ma[][];
int d[maxn];
int root;
bool sv(){
    queue<int>q;
    for(int i = ; i <= n; i++){
        if(d[i]==)q.push(i);
    }
    if(q.size()!=)return false;
    root = q.front();
    while(!q.empty()){
        int top = q.front();
        q.pop();
        for(int i = ; i < v[top].size(); i++){
            int x = v[top][i];
            d[x]--;
            if(f[x]!=-&&!ma[f[x]][top])return false;
            f[x] = top;
            if(d[x]==)q.push(x);
        }
    }

    for(int i = ; i <= n; i++){
        if(d[i]!=)return false;
    }f[root]=;
    for(int i = ; i <= n; i++){
        //if(i!=root){
            for(int j = ; j < v[i].size(); j++){
                if(!ma[f[i]][v[i][j]]){
                    //printf("--%d %d %d\n",i,f[i],v[i][j]);
                    return false;
                }
            }
        //}
    }
    return true;
}
int main() {
    scanf("%d", &n);
    for(int i = ; i <= n; i++){
        ma[][i] = ;
        f[i] = -;
        int k;
        scanf("%d", &k);
        for(int j = ; j <= k; j++){
            int x;
            scanf("%d", &x);
            v[i].pb(x);
            d[x]++;
            ma[i][x]=;
        }
    }
    mem(f,-);

    if(sv()){printf("YES\n");
    for(int i = ; i <= n; i++){
        if(i==root)continue;
        printf("%d %d\n",f[i],i);
    }}
    else printf("NO");
    return ;
}
/*
4
2 2 3
2 3 4
1 4
0
 */