无向图的边双连通分量

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无向图边双连通分量：对于一个无向图，任意两点之间至少存在两条边不重复的路径，则图是双连通的；边双连通的极大子图为边双连通分量；

求解边双连通分量：

1、dfs找出所有的桥；

2、把桥边删除，原图形成多个连通块，每个连通块就是一个边双连通分量；（只需再次dfs不经过桥就行）；

poj3352 求边双连通分量；

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<climits>#include<cctype>#include<iostream>#include<algorithm>#include<queue>#include<vector>#include<set>#include<stack>#include<string>#include<map>#define ll long long#define MAX 1010#define INF INT_MAX#define eps 1e-8using namespace std;struct Edge{int from,to;};vector<int>G[MAX];vector<Edge>edges;int pre[MAX],low[MAX],iscut[MAX][MAX],c[MAX],dfs_clock,n;void init(){for (int i = 0; i<=n; i++) G[i].clear();edges.clear();memset(iscut,0,sizeof(iscut));memset(pre,0,sizeof(pre));memset(c,0,sizeof(c));dfs_clock = 0;}int dfs(int u, int fa){int lowu = pre[u] = ++dfs_clock;for (int i=0; i<G[u].size(); i++){int v = G[u][i];if (!pre[v]){int lowv = dfs(v,u);lowu = min(lowu,lowv);if (lowv > pre[u]){       //标记桥iscut[u][v]  = iscut[v][u] = 1;edges.push_back((Edge){u,v});}}else if (pre[v] < pre[u] && v != fa){lowu = min(lowu,pre[v]);}}low[u] = lowu;return lowu;}int bcc[MAX],cnt;void dfs1(int u){       //求边双连通分量if (bcc[u]) return;bcc[u] = cnt;for(int i=0; i<G[u].size(); i++){int v = G[u][i];if (iscut[u][v]) continue;dfs1(v);}}int main(){int m;while (scanf("%d%d",&n,&m) != EOF){int u,v;init();for (int i=0; i<m; i++){scanf("%d%d",&u,&v);G[u].push_back(v);G[v].push_back(u);}dfs(1,-1);//for (int i=0; i<edges.size(); i++) printf("%d  %d\n",edges[i].from,edges[i].to);memset(bcc,0,sizeof(bcc));cnt = 0;for (int i=1; i<=n; i++){if (!bcc[i]){cnt++;dfs1(i);}}//printf("\n");//for (int i=1; i<=n; i++) printf("%d ",bcc[i]); printf("\n");for (int i=0; i<edges.size(); i++){int x = edges[i].from;int y = edges[i].to;c[bcc[x]]++;c[bcc[y]]++;}int ans = 0;for (int i=1; i<=cnt; i++) if (c[i] == 1) ans++;printf("%d\n",(ans+1) / 2);}return 0;}

这个题需要用到的命题：

一颗树有n（n>=2）个叶子节点的树，至少（只需）添加(n+1)/2条边，才（就）能转化为没有桥的图。或者说，使得图中的每条边都至少在一个环上；

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