#include <cstring> #include <iostream> #include <algorithm> #include <vector> using namespace std; typedef unsigned long long ULL; const int N = 5e5+10, M = 5e5+10; int n, m; int h[N], e[M], ne[M], idx; int dfn[N], low[N], timestamp; int stk[N], top; int dcc_cnt; //存每个双连通分量里有哪些点 vector<int> dcc[N]; //每个点是不是割点 bool cut[N]; //特判根节点 int root; int a,b; void add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ; } void tarjan(int u) { dfn[u] = low[u] = ++ timestamp; for (int i = h[u]; ~i; i = ne[i]) { int j = e[i]; //如果没被遍历过 if (!dfn[j]) { tarjan(j); low[u] = min(low[u], low[j]); // 表示b在j的子树中,就找到了 if (dfn[u] <= low[j]&&u!=a&&dfn[b]>=dfn[j]) { cut[u] = true; return ; } } else low[u] = min(low[u], dfn[j]); } } int main() { idx = n = timestamp = top = dcc_cnt = 0; memset(h, -1, sizeof h); memset(dfn, 0, sizeof dfn); memset(cut, 0, sizeof cut); cin>>n; while (cin>>a>>b) { if(a==0&&b==0) break; add(a, b), add(b, a); } cin>>a>>b; //以a为根 tarjan(a); for(int i=1; i<=n; i++) if(cut[i]) { cout<<i<<endl; return 0; } puts("No solution"); return 0; }