无向图点双联通。二分图判定
<span style="font-size:18px;">#include <cstdio> #include <stack> #include <vector> #include <algorithm> #include <cstring> using namespace std; struct Edge{ int u, v; }; const int maxn = 1005; int pre[maxn], iscut[maxn], bccno[maxn],dfs_clock, bcc_cnt; vector<int> G[maxn], bcc[maxn]; stack<Edge> S; int dfs(int u, int fa){ int lowu = pre[u] = ++dfs_clock; int child = 0; for(int i=0; i<G[u].size(); ++i){ int v = G[u][i]; Edge e = (Edge){u, v}; if(!pre[v]){ S.push(e); child++; int lowv = dfs(v, u); lowu = min(lowu, lowv); if(lowv>=pre[u]){ iscut[u] = true; bcc_cnt++; bcc[bcc_cnt].clear(); for(;;){ Edge x = S.top(); S.pop(); if(bccno[x.u]!=bcc_cnt){ bcc[bcc_cnt].push_back(x.u); bccno[x.u] = bcc_cnt; } if(bccno[x.v] != bcc_cnt){ bcc[bcc_cnt].push_back(x.v); bccno[x.v] = bcc_cnt; } if(x.u==u && x.v==v) break; } } } else if(pre[v]<pre[u] &&v!=fa){ S.push(e); lowu = min(lowu, pre[v]); } } if(fa < 0 && child == 1) iscut[u] = 0; return lowu; } void find_bcc(int n){ memset(pre, 0, sizeof pre ); memset(iscut, 0, sizeof iscut ); memset(bccno, 0, sizeof bccno ); dfs_clock = bcc_cnt = 0; for(int i=0; i<n; ++i) if(!pre[i]) dfs(i, -1); } int odd[maxn], color[maxn]; bool bipartite(int u, int b){ for(int i=0; i<G[u].size(); ++i){ int v = G[u][i]; if(bccno[v]!=b)continue; if(color[v]==color[u]) return false; if(!color[v]){ color[v] = 3 - color[u]; if(!bipartite(v, b)) return false; } } return true; } int A[maxn][maxn]; int main(){ #ifndef ONLINE_JUDGE freopen("in.cpp", "r", stdin); freopen("out.cpp", "w", stdout); #endif // ONLINE_JUDGE int kase = 0, n, m; while(scanf("%d%d", &n, &m)==2 &&n){ for(int i=0; i<n; ++i)G[i].clear(); memset(A, 0, sizeof A ); for(int i=0; i<m; ++i){ int u, v; scanf("%d%d", &u, &v); u--; v--; A[u][v] = A[v][u] = 1; } for(int u=0; u<n; ++u)//避免重边 for(int v=u+1; v<n; ++v) if(!A[u][v]) { G[u].push_back(v); G[v].push_back(u); } find_bcc(n); memset(odd, 0, sizeof odd ); for(int i=1; i<=bcc_cnt; ++i){ memset(color, 0, sizeof color ); for(int j=0; j<bcc[i].size(); ++j) bccno[bcc[i][j]] = i; int u = bcc[i][0]; color[u] = 1; if(!bipartite(u, i)) for(int j=0; j<bcc[i].size(); ++j) odd[bcc[i][j]] = 1; } int ans = n; for(int i=0; i<n; ++i) if(odd[i]) ans--; printf("%d ", ans); } return 0; } </span>