题目链接

https://atcoder.jp/contests/agc029/tasks/agc029_f

题解

考虑如何才能构成一棵树：显然有一个必要条件是对于每个点\(u\)来说，整张图所有的边与除去\(u\)之外所有的点存在完美匹配（即考虑一张二分图左边是除了\(u\)之外点的集合右边是\(E_i\), \(u\)和\(E_i\)连边当且仅当\(u\in E_i\)，该图存在完美匹配）。用Hall定理来表达就是，设\(S\)为\(\{ E_1,E_2,..,E_n\}\)的任意一个子集，\(N(S)\)表示这些\(E_i\)对应的点集的并集，则\(|N(S)|\ge |S|+1\).
实际上这个条件也是充分条件。我们用一个构造算法来证明。直接从\(1\)号点开始BFS或者DFS，每次\(u\)选择一条出边到右边的一个点\(v\)，然后跳到右边的点的匹配点\(u'\). 若这两个点都没被访问过，则添加一条树边\((u,u')\). 上面的命题等价于这样搜索能够遍历所有的点。因为如果某一个时刻不能走到未走过的点，那么走过的左边点个数比右边点个数多\(1\)，左边点总个数比右边点总个数多\(1\)，故现在未遍历的右边的点的集合\(T\)满足\(N(T)\le T\)，这与上面的命题矛盾。而如果上面的命题不成立，显然无法搜出合法的方案。而这样搜索能遍历所有的点等价于原问题有解，故原命题等价于原问题有解。
时间复杂度\(O(\sum |E_i|\sqrt n)\).

注: 在这里由于时间复杂度的限制，我们只能用二分图匹配检验\(1\)号点是否满足去掉后有完美匹配，但是这并不代表所有点都有。我们证明了有解的充要条件是每个点去掉后都有完美匹配，也是\(1\)号点去掉后有完美匹配且BFS/DFS不会在遍历完所有点前终止，我们的算法是正确的。但是似乎数据里并没有这种\(1\)号点去掉后有完美匹配但实际上无解的情况，把下面BFS的代码中标注comment_1的那一行return 0;前面加一个assert(0);依然可以AC. 但实际上这种情况完全可能出现，Hack数据如下：

4
4 1 2 3 4
2 3 4
2 3 4

代码

BFS

#include<bits/stdc++.h>
#define llong long long
#define mkpr make_pair
#define riterator reverse_iterator
#define pii pair<int,int>
using namespace std;inline int read()
{int x = 0,f = 1; char ch = getchar();for(;!isdigit(ch);ch=getchar()) {if(ch=='-') f = -1;}for(; isdigit(ch);ch=getchar()) {x = x*10+ch-48;}return x*f;
}const int INF = 1e7;namespace NetFlow
{const int N = 2e5+2;const int M = 4e5;struct Edge{int v,w,nxt,rev;} e[(M<<1)+3];int fe[N+3];int te[N+3];int dep[N+3];int que[N+3];int n,en,s,t;void addedge(int u,int v,int w){en++; e[en].v = v; e[en].w = w;e[en].nxt = fe[u]; fe[u] = en; e[en].rev = en+1;en++; e[en].v = u; e[en].w = 0;e[en].nxt = fe[v]; fe[v] = en; e[en].rev = en-1;}bool bfs(){for(int i=1; i<=n; i++) dep[i] = 0;int head = 1,tail = 1; que[1] = s; dep[s] = 1;while(head<=tail){int u = que[head]; head++;for(int i=fe[u]; i; i=e[i].nxt){int v = e[i].v;if(e[i].w>0 && dep[v]==0){dep[v] = dep[u]+1;if(v==t) return true;tail++; que[tail] = v;}}}return false;}int dfs(int u,int cur){if(u==t||cur==0) {return cur;}int rst = cur;for(int &i=te[u]; i; i=e[i].nxt){int v = e[i].v;if(e[i].w>0 && rst>0 && dep[v]==dep[u]+1){int flow = dfs(v,min(rst,e[i].w));if(flow>0){e[i].w -= flow; rst -= flow;e[e[i].rev].w += flow;if(rst==0) {return cur;}}}}if(rst==cur) {dep[u] = -2;}return cur-rst;}int dinic(int _n,int _s,int _t){n = _n,s = _s,t = _t;int ret = 0;while(bfs()){for(int i=1; i<=n; i++) te[i] = fe[i];memcpy(te,fe,sizeof(int)*(n+1));ret += dfs(s,INF);}return ret;}
}
using NetFlow::addedge;
using NetFlow::dinic;const int N = 1e5;
vector<int> adj[(N<<1)+3];
vector<pair<int,pii> > ans;
int mch[(N<<1)+3];
bool vis[(N<<1)+3];
int que[N+3];
int n;bool bfs()
{int hd = 1,tl = 1; que[1] = 1; vis[1] = true;while(hd<=tl){int u = que[hd]; hd++;for(int o=0; o<adj[u].size(); o++){int v = adj[u][o]; if(vis[v]) continue;if(vis[mch[v]]) continue;que[++tl] = mch[v]; vis[v] = vis[mch[v]] = true;ans.push_back(mkpr(v,mkpr(u,mch[v])));}}if(tl<n) {return false;}return true;
}int main()
{scanf("%d",&n);for(int i=1; i<=n; i++) addedge(1,i+2,1);for(int i=n+1; i<n+n; i++) addedge(i+2,2,1);for(int i=1; i<n; i++){int sz; scanf("%d",&sz);while(sz--){int x; scanf("%d",&x); adj[i+n].push_back(x); adj[x].push_back(i+n);if(x!=1) {addedge(x+2,i+n+2,1);}}}if(dinic(n+n+1,1,2)<n-1) {puts("-1"); return 0;}for(int u=3; u<=n+2; u++){for(int i=NetFlow::fe[u]; i; i=NetFlow::e[i].nxt){int v = NetFlow::e[i].v; if(v<=n+2) continue;if(NetFlow::e[i].w==0){mch[u-2] = v-2,mch[v-2] = u-2;break;}}}
//  printf("match: "); for(int i=1; i<=n+n-1; i++) printf("%d ",mch[i]); puts("");if(!bfs()) {puts("-1"); return 0;} //comment_1sort(ans.begin(),ans.end());for(int i=0; i<ans.size(); i++) printf("%d %d\n",ans[i].second.first,ans[i].second.second);return 0;
}

DFS

#include<bits/stdc++.h>
#define llong long long
#define mkpr make_pair
#define riterator reverse_iterator
#define pii pair<int,int>
using namespace std;inline int read()
{int x = 0,f = 1; char ch = getchar();for(;!isdigit(ch);ch=getchar()) {if(ch=='-') f = -1;}for(; isdigit(ch);ch=getchar()) {x = x*10+ch-48;}return x*f;
}const int INF = 1e7;namespace NetFlow
{const int N = 2e5+2;const int M = 4e5;struct Edge{int v,w,nxt,rev;} e[(M<<1)+3];int fe[N+3];int te[N+3];int dep[N+3];int que[N+3];int n,en,s,t;void addedge(int u,int v,int w){en++; e[en].v = v; e[en].w = w;e[en].nxt = fe[u]; fe[u] = en; e[en].rev = en+1;en++; e[en].v = u; e[en].w = 0;e[en].nxt = fe[v]; fe[v] = en; e[en].rev = en-1;}bool bfs(){for(int i=1; i<=n; i++) dep[i] = 0;int head = 1,tail = 1; que[1] = s; dep[s] = 1;while(head<=tail){int u = que[head]; head++;for(int i=fe[u]; i; i=e[i].nxt){int v = e[i].v;if(e[i].w>0 && dep[v]==0){dep[v] = dep[u]+1;if(v==t) return true;tail++; que[tail] = v;}}}return false;}int dfs(int u,int cur){if(u==t||cur==0) {return cur;}int rst = cur;for(int &i=te[u]; i; i=e[i].nxt){int v = e[i].v;if(e[i].w>0 && rst>0 && dep[v]==dep[u]+1){int flow = dfs(v,min(rst,e[i].w));if(flow>0){e[i].w -= flow; rst -= flow;e[e[i].rev].w += flow;if(rst==0) {return cur;}}}}if(rst==cur) {dep[u] = -2;}return cur-rst;}int dinic(int _n,int _s,int _t){n = _n,s = _s,t = _t;int ret = 0;while(bfs()){for(int i=1; i<=n; i++) te[i] = fe[i];memcpy(te,fe,sizeof(int)*(n+1));ret += dfs(s,INF);}return ret;}
}
using NetFlow::addedge;
using NetFlow::dinic;const int N = 1e5;
vector<int> adj[(N<<1)+3];
vector<pair<int,pii> > ans;
int mch[(N<<1)+3];
bool vis[(N<<1)+3];
int n;void dfs(int u)
{for(int o=0; o<adj[u].size(); o++){int v = adj[u][o]; if(vis[v]) continue;if(vis[mch[v]]) continue;vis[v] = vis[mch[v]] = true; ans.push_back(mkpr(v,mkpr(u,mch[v])));dfs(mch[v]);}
}int main()
{scanf("%d",&n);for(int i=1; i<=n; i++) addedge(1,i+2,1);for(int i=n+1; i<n+n; i++) addedge(i+2,2,1);for(int i=1; i<n; i++){int sz; scanf("%d",&sz);while(sz--){int x; scanf("%d",&x); adj[i+n].push_back(x); adj[x].push_back(i+n);if(x!=1) {addedge(x+2,i+n+2,1);}}}if(dinic(n+n+1,1,2)<n-1) {puts("-1"); return 0;}for(int u=3; u<=n+2; u++){for(int i=NetFlow::fe[u]; i; i=NetFlow::e[i].nxt){int v = NetFlow::e[i].v; if(v<=n+2) continue;if(NetFlow::e[i].w==0){mch[u-2] = v-2,mch[v-2] = u-2;break;}}}
//  printf("match: "); for(int i=1; i<=n+n-1; i++) printf("%d ",mch[i]); puts("");vis[1] = true; dfs(1);if(ans.size()<n-1) {puts("-1"); return 0;}sort(ans.begin(),ans.end());for(int i=0; i<ans.size(); i++) printf("%d %d\n",ans[i].second.first,ans[i].second.second);return 0;
}