UVa

UVa01660 Cable TV Network

題目連結：https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=0&problem=4535&mosmsg=Submission+received+with+ID+24585550
題意：給一張圖，求點連通度，即為需要拔掉幾個點，才能讓圖不連通。

解法：如果求邊連通度，直接枚舉點對用最大流去算，點連通度的話，要把點成兩個，之間流量為 $1$，原圖上的邊流量設無限大，之後一樣枚舉點對用最大流去算出(可以一個匯點，枚舉其他點為匯點，就能算出答案)。

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#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;
const int INF = 1e9;
const int MXN = 0;
const int MXV = 150;
const int MXE = 3000;
#define MP make_pair
#define PB push_back
#define F first
#define S second
#define FOR(i, L, R) for (int i = L; i != (int)R; ++i)
#define FORD(i, L, R) for (int i = L; i != (int)R; --i)
#define IOS                                                                    \
    cin.tie(nullptr);                                                          \
    cout.tie(nullptr);                                                         \
    ios_base::sync_with_stdio(false);

template <typename T> struct Dinic
{
    int n, s, t, level[MXV], now[MXV];
    struct Edge
    {
        int v;
        T rf; // rf: residual flow
        int re;
    };
    vector<Edge> e[MXV];
    void init(int _n, int _s, int _t)
    {
        n = _n;
        s = _s;
        t = _t;
        for (int i = 0; i <= n; i++)
        {
            e[i].clear();
        }
    }
    void add_edge(int u, int v, T f)
    {
        // cout << u << ' ' << v << ' ' << f << '\n';
        e[u].push_back({v, f, (int)e[v].size()});
        e[v].push_back({u, 0, (int)e[u].size() - 1});
    }
    bool bfs()
    {
        fill(level, level + n + 1, -1);
        queue<int> q;
        q.push(s);
        level[s] = 0;
        while (!q.empty())
        {
            int u = q.front();
            q.pop();
            for (auto it : e[u])
            {
                if (it.rf > 0 && level[it.v] == -1)
                {
                    level[it.v] = level[u] + 1;
                    q.push(it.v);
                }
            }
        }
        return level[t] != -1;
    }
    T dfs(int u, T limit)
    {
        // cout << u << ' ' << limit << '\n';
        if (u == t)
        {
            return limit;
        }
        T res = 0;
        while (now[u] < (int)e[u].size())
        {
            Edge &it = e[u][now[u]];
            if (it.rf > 0 && level[it.v] == level[u] + 1)
            {
                T f = dfs(it.v, min(limit, it.rf));
                res += f;
                limit -= f;
                it.rf -= f;
                e[it.v][it.re].rf += f;
                if (limit == 0)
                {
                    return res;
                }
            }
            else
            {
                ++now[u];
            }
        }
        if (!res)
        {
            level[u] = -1;
        }
        return res;
    }
    T flow(T res = 0)
    {
        while (bfs())
        {
            T tmp;
            memset(now, 0, sizeof(now));
            do
            {
                tmp = dfs(s, INF);
                res += tmp;
            } while (tmp);
        }
        // cout << s << ' ' << t << ' ' << res << '\n';
        return res;
    }
};

int main()
{
    // IOS;
    int n, m;
    vector<int> u, v;
    Dinic<int> dinic;
    while (cin >> n >> m)
    {
        // cout << n << ' ' << m << '\n';
        int ans = INF;
        char ch;
        u.resize(m + 5);
        v.resize(m + 5);
        FOR(i, 0, m) {
            cin >> ch >> u[i] >> ch >> v[i] >> ch;
        }
        FOR(s, 0, n)
        {
            FOR(t, s + 1, n)
            {
                // cout << s << ' ' << t << '\n';
                dinic.init(2 * n, s + n, t);
                FOR(i, 0, n) { dinic.add_edge(i, i + n, 1); }
                FOR(i, 0, m)
                {
                    dinic.add_edge(u[i] + n, v[i], INF);
                    dinic.add_edge(v[i] + n, u[i], INF);
                }
                ans = min(ans, dinic.flow());
            }
        }
        if (ans == INF)
        {
            cout << n << '\n';
        }
        else
        {
            cout << ans << '\n';
        }
    }
}

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