BZOJ 1093. [ZJOI2007]最大半连通子图

如果这张图是个DAG，那么最长链就是第一个答案，所以就先tarjan缩点。

第二部分拓扑排序解决，注意重边会影响答案，所以不要重复转移。

#include <bits/stdc++.h>

namespace IO {
    char buf[1 << 21], *p1 = buf, *p2 = buf;
    int p, p3 = -1;
    void read() {}
    inline int getc() {
        return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
    }
    template <typename T, typename... T2>
    inline void read(T &x, T2 &... oth) {
        T f = 1; x = 0;
        char ch = getc();
        while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getc(); }
        while (isdigit(ch)) { x = x * 10 + ch - 48; ch = getc(); }
        x *= f;
        read(oth...);
    }
}

const int N = 2e5 + 7;
const int M = 2e6 + 7;
struct E {
    int v, ne;
} e1[M], e2[M];
int head1[N], head2[N], cnt1 = 1, cnt2 = 1;
int n, m, MOD, dfn[N], low[N], st[N], top, tol;
int belong[N], num, sz[N], d[N];
bool in[N];

void tarjan(int u) {
    dfn[u] = low[u] = ++tol;
    st[++top] = u;
    in[u] = 1;
    for (int i = head1[u]; i; i = e1[i].ne) {
        int v = e1[i].v;
        if (!dfn[v])
            tarjan(v), low[u] = std::min(low[u], low[v]);
        else if (in[v])
            low[u] = std::min(low[u], dfn[v]);
    }
    if (low[u] == dfn[u]) {
        ++num;
        int v;
        do {
            v = st[top--];
            belong[v] = num;
            sz[num]++;
            in[v] = 0;
        } while (u != v);
    }
}

struct Edge {
    int u, v;
} Edge[M];

void add1(int u, int v) {
    e1[++cnt1].v = v; e1[cnt1].ne = head1[u]; head1[u] = cnt1;
}

void add2(int u, int v) {
    e2[++cnt2].v = v; e2[cnt2].ne = head2[u]; head2[u] = cnt2;
}

struct Node {
    int size, cnt;
    Node(int s = 0, int c = 0): size(s), cnt(c) {}
    Node operator + (int a) const {
        return Node(size + a, cnt);
    }
    Node operator * (const Node &p) const {
        if (size > p.size) return *this;
        if (size < p.size) return p;
        return Node(size, (cnt + p.cnt) % MOD);
    }
} dp[N], ans;
int last[N];

int main() {
    IO::read(n, m, MOD);
    for (int i = 1; i <= m; i++) {
        IO::read(Edge[i].u, Edge[i].v);
        add1(Edge[i].u, Edge[i].v);
    }
    for (int i = 1; i <= n; i++)
        if (!dfn[i])
            tarjan(i);
    for (int i = 1; i <= n; i++) {
        for (int j = head1[i]; j; j = e1[j].ne) {
            int u = belong[i], v = belong[e1[j].v];
            if (u != v)
                add2(u, v), d[v]++;
        }
    }
    std::queue<int> que;
    for (int i = 1; i <= num; i++)
        if (!d[i])
            que.push(i), dp[i] = Node(sz[i], 1);
    while (!que.empty()) {
        int u = que.front(); que.pop();
        ans = ans * dp[u];
        for (int i = head2[u]; i; i = e2[i].ne) {
            int v = e2[i].v;
            if (last[v] != u) {
                last[v] = u;
                dp[v] = dp[v] * (dp[u] + sz[v]);
            }
            --d[v];
            if (!d[v])
                que.push(v);
        }
    }
    printf("%d\n%d\n", ans.size, ans.cnt);
    return 0;
}

标签：Node,ch,1093,int,子图,++,low,ZJOI2007,size
来源： https://www.cnblogs.com/Mrzdtz220/p/12234377.html