Redundant Paths 分离的路径

#include<iostream>
#include<stack>
#include<cstdio>
using namespace std;
const int N = 12000;

struct edge{
    int to;
    int next;
}e[N];

int n,m;
int tot;
int head[N];
int dfn[N];
int low[N];
int id[N];
int t;
int cnt;
int ru[N];
int ans;

stack <int> st;

void add(int x,int y){
    tot++;
    e[tot].to = y;
    e[tot].next = head[x];
    head[x] = tot;
}

void tarjan(int x,int ok){
    low[x] = dfn[x] = ++t;
    st.push(x);
    for(int i=head[x];i;i=e[i].next){
        int y = e[i].to;
        if(!dfn[y]){
            tarjan(y,i);
            low[x] = min(low[x],low[y]);
        }
        else if(i != (ok^1))
            low[x] = min(low[x],dfn[y]); 
    }
    if(low[x] == dfn[x]){
        int k;
        cnt++;
        while(k != x){
            k = st.top();
            st.pop();
            id[k] = cnt;
        }
    }
}

int main(){
    scanf("%d%d",&n,&m);
    tot=1;
    for(int i=1;i<=m;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        add(x,y);
        add(y,x);
    }
    for(int i=1;i<=n;i++)
        if(!dfn[i])tarjan(i,-1);
    for(int x=1;x<=n;x++){
        for(int i=head[x];i;i=e[i].next){
            int y = e[i].to;
            if(id[x] == id[y]) continue;
            ru[id[y]]++;
        }
    }
    for(int i=1;i<=cnt;i++)
        if(ru[i] == 1)ans++;
    ans = (ans + 1)/2;
    printf("%d
",ans);
    return 0;
}