題目鏈接

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

題解

考慮如何才能構(gòu)成一棵樹：顯然有一個必要條件是對于每個點\(u\)來說，整張圖所有的邊與除去\(u\)之外所有的點存在完美匹配（即考慮一張二分圖左邊是除了\(u\)之外點的集合右邊是\(E_i\), \(u\)和\(E_i\)連邊當且僅當\(u\in E_i\)，該圖存在完美匹配）。用Hall定理來表達就是，設(shè)\(S\)為\(\{ E_1,E_2,..,E_n\}\)的任意一個子集，\(N(S)\)表示這些\(E_i\)對應(yīng)的點集的并集，則\(|N(S)|\ge |S|+1\).
實際上這個條件也是充分條件。我們用一個構(gòu)造算法來證明。直接從\(1\)號點開始BFS或者DFS，每次\(u\)選擇一條出邊到右邊的一個點\(v\)，然后跳到右邊的點的匹配點\(u'\). 若這兩個點都沒被訪問過，則添加一條樹邊\((u,u')\). 上面的命題等價于這樣搜索能夠遍歷所有的點。因為如果某一個時刻不能走到未走過的點，那么走過的左邊點個數(shù)比右邊點個數(shù)多\(1\)，左邊點總個數(shù)比右邊點總個數(shù)多\(1\)，故現(xiàn)在未遍歷的右邊的點的集合\(T\)滿足\(N(T)\le T\)，這與上面的命題矛盾。而如果上面的命題不成立，顯然無法搜出合法的方案。而這樣搜索能遍歷所有的點等價于原問題有解，故原命題等價于原問題有解。
時間復雜度\(O(\sum |E_i|\sqrt n)\).

注: 在這里由于時間復雜度的限制，我們只能用二分圖匹配檢驗\(1\)號點是否滿足去掉后有完美匹配，但是這并不代表所有點都有。我們證明了有解的充要條件是每個點去掉后都有完美匹配，也是\(1\)號點去掉后有完美匹配且BFS/DFS不會在遍歷完所有點前終止，我們的算法是正確的。但是似乎數(shù)據(jù)里并沒有這種\(1\)號點去掉后有完美匹配但實際上無解的情況，把下面BFS的代碼中標注comment_1的那一行return 0;前面加一個assert(0);依然可以AC. 但實際上這種情況完全可能出現(xiàn)，Hack數(shù)據(jù)如下：

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

代碼

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; }

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