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(uin 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. 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_1
sort(ans.begin(),ans.end());
for(int i=0; i<ans.size(); i++) printf("%d %d
",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
",ans[i].second.first,ans[i].second.second);
return 0;
}