求割点与桥
<pre><code class="language-cpp">#include<iostream>
using namespace std;
#include<cstdio>
#include<cstring>
#include<vector>
const int maxn = 1e5+5;
vector<int> G[maxn];
int n, m, low[maxn], dfn[maxn];
bool is_cut[maxn];
int father[maxn];
int tim = 0;
void input(){
scanf("%d%d",&n, &m);
int a, b;
for(int i = 1; i <= m; ++i){
scanf("%d%d",&a, &b);
G[a].push_back(b);/*邻接表储存无向边*/
G[b].push_back(a);
}
}
void Tarjan(int i, int Father){
father[i] = Father;/*记录每一个点的父亲*/
dfn[i] = low[i] = tim++;
for(int j = 0; j < G[i].size(); ++j){
int k = G[i][j];
if(dfn[k] == -1){
Tarjan(k, i);
low[i] = min(low[i], low[k]);
}
else if(Father !=k)/*假如k是i的父亲的话，那么这就是无向边中的重边，有重边那么一定不是桥*/
low[i] = min(low[i], dfn[k]);//dfn[k]可能！=low[k]，所以不能用low[k]代替dfn[k],否则会上翻过头了。
}
}
void count(){
int rootson = 0;
Tarjan(1, 0);
for(int i = 2; i <= n; ++i){
int v = father[i];
if(v == 1)
rootson++;/*统计根节点子树的个数，根节点的子树个数>=2,就是割点*/
else{
if(low[i] >= dfn[v])/*割点的条件*/
is_cut[v] = true;//这里可以改成计数，来判断去掉后产生多少个连通过快
}
}
if(rootson > 1) is_cut[1] = true;
for(int i = 1; i <= n; ++i){
if(is_cut[i]) printf("%d\n",i);//输出割点
}
for(int i = 1; i <= n; ++i){
int v = father[i];
if(v > 0 && low[i] > dfn[v])/*桥的条件*/
printf("%d,%d\n",v,i);
}
}
int main(){
input();
memset(dfn,-1,sizeof(dfn));
memset(father,0,sizeof(father));
memset(low,-1,sizeof(low));
memset(is_cut,false,sizeof(is_cut));
count();
return 0;
}</code></pre>