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冗余路径「边双连通分量、无向图缩点」

冗余路径

题目描述:

思路:

#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define inf 0x3f3f3f3f
#define mod7 1000000007
#define mod9 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug(a) cout << "Debuging...|" << #a << ": " << a << "\n";
typedef long long ll;
typedef pair <int,int> pii;

#define MAX 300000 + 50
int n, m, k;

int tot;
int head[MAX];
struct ran{
    int to, nex;
}tr[MAX];
inline void add(int u, int v){
    tr[++tot].to = v;
    tr[tot].nex = head[u];
    head[u] = tot;
}

int tmd;
int dfn[MAX], low[MAX];
bool vis[MAX];
stack<int>st;
int cnt;
int col[MAX];
int du[MAX];
bool g[MAX];
void tarjan(int u){
    st.push(u);vis[u] = 1;
    dfn[u] = low[u] = ++tmd;
    for(int i = head[u]; i; i = tr[i].nex){
        if(g[i])continue;//判重边
        g[i] = g[i ^ 1] = 1;
        int v = tr[i].to;
        if(!dfn[v]){
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(vis[v])low[u] = min(low[u], dfn[v]);
    }
    if(low[u] == dfn[u]){
        ++cnt;
        int now;
        do {
            now = st.top();
            st.pop();vis[now] = 0;
            col[now] = cnt;
        } while (now != u);
    }
}

void work(){
    cin >> n >> m;
    tot = 1;
    for(int i = 1, a, b; i <= m; ++i){
        cin >> a >> b;
        add(a, b);
        add(b, a);
    }
    for(int i = 1; i <= n; ++i){
        if(!dfn[i])tarjan(i);
    }
    for(int u = 1; u <= n; ++u){
        for(int i = head[u]; i; i = tr[i].nex){
            int v = tr[i].to;
            if(col[u] != col[v]){
                ++du[col[v]];
            }
        }
    }
    int ans = 0;
    for(int i = 1; i <= cnt; ++i){
        if(du[i] == 1)++ans;
    }
    cout << (ans + 1) / 2 << endl;
    
}


int main(){
    io;
    work();
    return 0;
}
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