P2746 [USACO5.3]校园网Network of Schools tarjan 缩点

题意

给出一个有向图，
A任务：求最少需要从几个点送入信息，使得信息可以通过有向图走遍每一个点
B任务：求最少需要加入几条边，使得有向图是一个强联通分量

思路

任务A，比较好想，可以通过tarjan缩点，求出入度为0的点的个数
任务B
一开始以为任务A，B没有关系
其实是入度为0的点的个数、出度为0的点的个数的最大值。
因为任务B要求在任意学校投放软件使得所有学校都能收到，所以很明显是需要整张图形成一个环，而环中所有节点入度和出度都不为0，所以需要把所有入度和出度的点度数增加。

（注意判断本身就全联通的情况

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}

/*-----------------------showtime----------------------*/

const int maxn = 109;
vector<int> mp1[maxn], mp2[maxn];

int dfn[maxn], low[maxn], vis[maxn], col[maxn];
int dp[maxn][maxn], in[maxn];
stack<int> st;
int tot = 0, nn = 0;
void tarjan(int u) {
    dfn[u] = low[u] = ++tot;

    st.push(u);
    vis[u] = 1;
    for (int i = 0; i < mp1[u].size(); i++) {
        int v = mp1[u][i];
        if (dfn[v] == 0) {
            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]) {
        nn++;
        while (!st.empty()) {
            int x = st.top();
            st.pop();
            col[x] = nn;
            vis[x] = 0;
            if (x == u) break;
        }
    }
}

int main() {
    int n;
    scanf("%d", &n);
    rep(i, 1, n) {
        int x;

        while (~scanf("%d", &x) && x) mp1[i].pb(x);
    }

    rep(i, 1, n) if (!dfn[i]) tarjan(i);
    rep(i, 1, n) {
        int u = col[i];
        for (int j = 0; j < mp1[i].size(); j++) {
            int v = col[mp1[i][j]];
            dp[u][v] = 1;
        }
    }
    rep(i, 1, nn) {
        rep(j, 1, nn) {
            if (i == j) continue;
            if (dp[i][j]) mp2[i].pb(j), in[j]++;
        }
    }

    int ansa = 0, c1 = 0, c2 = 0;
    rep(i, 1, nn) {
        if (in[i] == 0) ansa++, c1++;
        if (mp2[i].size() == 0) c2++;
    }
    if (nn == 1) c1 = c2 = 0;
    printf("%d\n%d\n", ansa, max(c1, c2));
    return 0;
}