题面

题目链接

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备用题面

给定一个 $N$ 个顶点，$M$ 条边的无向连通图，顶点从 $1$ 到 $N$ 编号。

问至少要添加几条边，才能使其变为双连通图。

思路

这道题目不难想到缩点以后，找到度为 $1$ 的点（因为这样一切就断了）然后这样点的个数我们要把他们连到一起，答案就是这样点的数量除以二去上底。

代码

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#pragma GCC optimize(2)
#pragma GCC optimize(3, "Ofast", "inline")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef __int128 int128;

namespace FastIO 
{
    char buf[1 << 20], *p1, *p2;
    #define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2) ? EOF : *p1++)
    template<typename T> inline T read(T& x) {
        x = 0;
        int f = 1;
        char ch;
        while (!isdigit(ch = getchar())) if (ch == '-') f = -1;
        while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
        x *= f;
        return x;
    }
    template<typename T, typename... Args> inline void read(T& x, Args &...x_) {
        read(x);
        read(x_...);
        return;
    }
    inline ll read() {
        ll x;
        read(x);
        return x;
    }
    inline void read(char *s) {
        scanf("%s", s + 1);
    }
};
using namespace FastIO;

const int N = 1010;
const int M = 2010;

class Graph {
    private :
        struct Edge {
            int to, nt, wt;
            Edge() {}
            Edge(int to, int nt, int wt) : to(to), nt(nt), wt(wt) {}
        }e[M];
        int hd[N], cnte;
    public :
        void AddEdge(int u, int v, int w = 0) {
            e[++cnte] = Edge(v, hd[u], w);
            hd[u] = cnte;
        }
        int head(int u) { return hd[u]; }
        int nt(int u) { return e[u].nt; }
        int to(int u) { return e[u].to; }
        int wt(int u) { return e[u].wt; }
};

int n, m;
Graph e;
int uid[M], vid[M];

inline void Input() {
    read(n, m);
    int u, v;
    for(int i = 1; i <= m; i++) {
        read(uid[i], vid[i]);
        e.AddEdge(uid[i], vid[i]);
        e.AddEdge(vid[i], uid[i]);
    }
}

int low[N], dfn[N], cntd;
int vis[N];
stack<int >st;
int cl[N], cntc;

int inD[N];

inline void Tarjan(int u, int fa) {
    vis[u] = 1;
    low[u] = dfn[u] = ++cntd;
    st.push(u);
    for(int i = e.head(u); i ; i = e.nt(i)) {
        int v = e.to(i);
        if(v == fa) continue;
        if(!dfn[v]) {
            Tarjan(v, u); 
            low[u] = min(low[u], low[v]);
        } else if(vis[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(dfn[u] == low[u]){
        int v = 0;
        ++cntc;
        while(u != v) {
            v = st.top(); st.pop();
            cl[v] = cntc;
            vis[v] = 0;
        }
    }
}

inline void Work() {
    Tarjan(1, 1);
    for(int i = 1; i <= m; i++) {
        if(cl[uid[i]] != cl[vid[i]]) {
            // cout << uid[i] << ' ' << vid[i] << endl;
            inD[cl[uid[i]]]++;
            inD[cl[vid[i]]]++;
        }
    }
    int ans = 0;
    for(int i = 1; i <= cntc; i++) {
        // cout << inD[i] << ' ';
        if(inD[i] == 1) ans++;
    }
    // cout <<endl;
    printf("%d", ans + 1 >> 1);
}

int main() {
    int T = 1;
    while(T--) {
        Input();
        Work();
    }
    return 0;
}