POJ 3177 Redundant Paths (tarjan无向图求缩点)

#include <iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<cstdio>
#include<stack>
#include <map>
#define lson(p) (p<<1)
#define rson(p) (p<<1|1)
#define ll long long
using namespace std;
const int N = 1e5+10;
int n,m;
struct Edge{
    int to,nxt;
}edge[N<<2];
int tot,head[N];
int dfn[N],low[N],fa[N],d[N];
stack<int> st;
int ind,ans;
void init(){
    tot = 0;
    memset(head,-1,sizeof head);
    memset(dfn,0,sizeof dfn);
    memset(low,0,sizeof low);
}
void add(int u,int v){
    edge[tot].to = v;
    edge[tot].nxt = head[u];
    head[u] = tot++;
}
void tarjan(int u,int root){
    int v,sg=0;
    dfn[u] = low[u] = ++ind;
    st.push(u);
    for(int i =head[u];i!=-1;i=edge[i].nxt){
        v = edge[i].to;
        if(v==root && !sg) {
            sg = 1;
            continue;
        }
        if(!dfn[v]){
            tarjan(v,u);
            low[u] = min(low[u],low[v]);
        }
        else if(dfn[v] < dfn[u]){
            low[u] = min(low[u],dfn[v]);
        }
    }
    if(low[u]== dfn[u]){
        int v = st.top();
        while(v !=u){
            fa[v] = u;
            st.pop();
            v = st.top();
        }
        fa[u] = u;
        st.pop();
    }
}
void solve(){
    int fr,to;
    init();
    for(int i=1;i<=m;++i){
        scanf("%d%d",&fr,&to);
        add(fr,to);
        add(to,fr);
    }
    tarjan(1,0);
    for(int i=1;i<=n;++i){
        for(int j=head[i];j!=-1;j=edge[j].nxt){
            if(fa[i]!=fa[edge[j].to]){
                d[fa[i]]++;
                d[fa[edge[j].to]]++;
            }
        }
    }
    int cnt = 1;
    for(int i=1;i<=n;++i){
        if(d[i]==2){
            cnt++;
        }
    }
    printf("%d\n",(cnt)/2);
}
int main()
{
    while(scanf("%d%d",&n,&m)){
        solve();
        break;
    }
    return 0;
}

来源：https://www.cnblogs.com/xxrlz/p/11368106.html