点连通分量 - 缩点

矿场搭建

\(\href{https://www.acwing.com/solution/content/24931/}{点连通分量的缩点}\)

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define ull unsigned long long
#define pb push_back
#define PII pair<int, int>
#define x first
#define y second
#define inf 0x3f3f3f3f
const int N = 1010, M = 1010;
int n, m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;
int stk[N], top;
int dcc_cnt;
vector<int> dcc[N];
bool cut[N];
int root;

void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

void tarjan(int u) {
    dfn[u] = low[u] = ++timestamp;
    stk[++top] = u;
    if (u == root && h[u] == -1) {
        dcc_cnt++;
        dcc[dcc_cnt].pb(u);
        return;
    }
    int cnt = 0;
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (!dfn[j]) {
            tarjan(j);
            low[u] = min(low[u], low[j]);
            if (dfn[u] <= low[j]) {
                cnt++;
                if (u != root || cnt > 1) cut[u] = 1;
                ++dcc_cnt;
                int y;
                do {
                    y = stk[top--];
                    dcc[dcc_cnt].pb(y);
                } while (y != j);
                dcc[dcc_cnt].pb(u);
            }
        }
        else low[u] = min(low[u], dfn[j]);
    }
}

int main() {
    IO;
    int T = 1;
    while (cin >> m, m) {
        for (int i = 1; i <= dcc_cnt; ++i) dcc[i].clear();
        idx = n = timestamp = top = dcc_cnt = 0;
        memset(h, -1, sizeof h);
        memset(dfn, 0, sizeof dfn);
        memset(cut, 0, sizeof cut);
        while (m--) {
            int a, b;
            cin >> a >> b;
            n = max(a, n), n = max(n, b);
            add(a, b), add(b, a);
        }
        for (root = 1; root <= n; ++root) 
            if (!dfn[root])
                tarjan(root);
        int res = 0;
        ull num = 1;
        for (int i = 1; i <= dcc_cnt; ++i) {
            int cnt = 0;
            for (int j = 0; j < dcc[i].size(); ++j)
                if (cut[dcc[i][j]]) cnt++;
            if (cnt == 0) {
                if (dcc[i].size() > 1) res += 2, num *= dcc[i].size() * (dcc[i].size() - 1) / 2;
                else res++;
            } else if (cnt == 1) res++, num *= dcc[i].size() - 1;
        }
        cout << "Case " << T++ << ": " <<  res << " " << num << '\n';
    }
    return 0;
}

标签：缩点,连通,int,cnt,dcc,++,low,define,分量
来源： https://www.cnblogs.com/phr2000/p/14427144.html