最大半连通子图

https://www.acwing.com/problem/content/1177/

\(强连通必然半连通,将强连通图转化为拓补图,从起点出发,能走的所有\color{Red}{最长的路径即为最大半连通子图}\)

\(问题就转化为在拓补图上做dp的问题\)

\(注意缩点后,每个点间只用保留一条边,连通子图只和点以及它们之间是否有连接关系有关,与有几条边无关.\)

#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 pb push_back
#define PII pair<int, int>
#define x first
#define y second
#define inf 0x3f3f3f3f
const int N = 1e5 + 10, M = 2e6 + 10;
int n, m, mod;
int h[N], hs[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;
int stk[N], top;
bool in_stk[N];
int id[N], scc_cnt, scc_size[N];
int f[N], g[N];

void add(int h[], 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, in_stk[u] = true;
    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]);
        }
        else if (in_stk[j]) low[u] = min(low[u], dfn[j]);
    }
    if (dfn[u] == low[u]) {
        ++scc_cnt;
        int y;
        do {
            y = stk[top--];
            in_stk[y] = false;
            id[y] = scc_cnt;
            scc_size[scc_cnt]++;
        } while (y != u);
    }
}

int main() {
    IO;
    cin >> n >> m >> mod;
    memset(h, -1, sizeof h);
    memset(hs, -1, sizeof hs);
    while (m--) {
        int a, b;
        cin >> a >> b;
        add(h, a, b);
    }
    for (int i = 1; i <= n; ++i) 
        if (!dfn[i]) tarjan(i);

    unordered_set<ll> S;
    for (int i = 1; i <= n; ++i)
        for (int j = h[i]; ~j; j = ne[j]) {
            int k = e[j];
            int a = id[i], b = id[k];
            ll hash = a * 1000000ll + b;
            if (a != b && !S.count(hash)) {
                add(hs, a, b);
                S.insert(hash);
            }
        }
    for (int i = scc_cnt; i; --i) {
        if (!f[i]) {
            f[i] = scc_size[i];
            g[i] = 1;
        }
        for (int j = hs[i]; ~j; j = ne[j]) {
            int k = e[j];
            if (f[k] < f[i] + scc_size[k]) {
                f[k] = f[i] + scc_size[k];
                g[k] = g[i];
            }
            else if (f[k] == f[i] + scc_size[k])
                g[k] = (g[k] + g[i]) % mod;
        }
    }
    
    int maxf = 0, sum = 0;
    for (int i = 1; i <= scc_cnt; ++i)
        if (f[i] > maxf) {
            maxf = f[i];
            sum = g[i];
        }
        else if (f[i] == maxf) sum = (sum + g[i]) % mod;
    cout << maxf << '\n';
    cout << sum << '\n';
    return 0;
}

posted @ 2021-02-20 01:50  phr2000  阅读(63)  评论(0编辑  收藏  举报