P2746 [USACO5.3] 校园网Network of Schools 题解

题目链接：校园网Network of Schools

这个题得翻译下题目意思才知道在干嘛，题目一开始表明了这个是一个有向图，因为边是单向的。其次关于第一个问题：

基于一个事实，如果有 \(x \rightarrow y \rightarrow z\)，那么只需要 \(x\) 接受协议，它所在的 \(scc\) 强连通分量上的点一定都能不需要接受协议了，那么其实做缩点以后，算入度为 \(0\) 的强联通分量数量就行了。

第二个问题，问还需要加多少条边完成互通，显然加到整体变成一个强连通分量就可以互通了。那么最少加几条边呢？如果原来本身就是强连通分量，那么显然不需要加边。否则考虑缩点以后：

如图所示，我们需要干一件事：

添加边连向出度为 \(0\) 的点与入度为 \(0\) 的点，使得它俩连通，至于怎么连无所谓。假设入度为 \(0\) 的点为 \(x\)，出度为 \(0\) 的点为 \(y\) 个。我们需要把所有的 \(x\) 和所有的 \(y\) 连接，连接方式随意，显然至少需要连 \(\max(x,y)\) 条边，当然如果本身是 \(scc\) 显然不需要连边，特判就行。至于求 \(scc\) 直接跑 \(tarjan\) 就行了，然后缩点，连边并不需要真的连，直接记录出入度变化就行了。

参照代码

#include <bits/stdc++.h>

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

#define isPbdsFile

#ifdef isPbdsFile

#include <bits/extc++.h>

#else

#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>

#endif

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}

template <typename T>
T lowBit(T x)
{
    return x & -x;
}

template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}

template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}

template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}

template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}

template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}

template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}

template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}


template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;

    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }

    T3() { one = tow = three = 0; }

    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};

template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}

template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}

constexpr int N = 110;
vector<int> child[N];
int scc[N], sccCnt;
int dfn[N], low[N];
int n, cnt;
stack<int> st;

inline void tarjan(const int curr)
{
    st.push(curr);
    dfn[curr] = low[curr] = ++cnt;
    for (const int nxt : child[curr])
    {
        if (!dfn[nxt])tarjan(nxt), uMin(low[curr], low[nxt]);
        else if (!scc[nxt])uMin(low[curr], dfn[nxt]);
    }
    if (dfn[curr] == low[curr])
    {
        ++sccCnt;
        while (true)
        {
            const int nxt = st.top();
            st.pop();
            scc[nxt] = sccCnt;
            if (nxt == curr)break;
        }
    }
}

int in[N], out[N];

inline void solve()
{
    cin >> n;
    forn(i, 1, n)
    {
        int son;
        cin >> son;
        while (son)child[i].push_back(son), cin >> son;
    }
    forn(i, 1, n)if (!dfn[i])tarjan(i);
    forn(i, 1, n)
    {
        for (const int j : child[i])
        {
            if (scc[i] != scc[j])
            {
                //scc[i]->scc[j]
                in[scc[j]]++;
                out[scc[i]]++;
            }
        }
    }
    int ansA = 0, ansB = 0;
    forn(i, 1, sccCnt)ansA += in[i] == 0, ansB += out[i] == 0;
    uMax(ansB, ansA);
    if (sccCnt == 1)ansB = 0;
    cout << ansA << endl << ansB;
}

signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

\[时间复杂度为:\ O(n+m),m为总边数 \]