acwing算法提高之图论--有向图的强连通分量

1 介绍

本博客介绍有向图的强连通分量的题目。

连通分量：是针对有向图的一个概念。对于分量中任意两个结点a、b，必然可以从a走到b，且从b走到a。
强连通分量：是针对有向图的一个概念。极大强连通分量，也就是说再加一个结点，它就不是连通分量。

强连通分量，用来将一个有向图转化为一个有向无环图（DAG、拓扑图）。方法是缩点，将所有连通分量缩成一个点。
有向无环图有很多好处，可以递推（即拓扑序）求最短路或最长路。

2 训练

题目1：1174受欢迎的牛

C++代码如下，

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 10010, M = 50010;

int n, m;
int h[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, Size[N];
int dout[N];

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, in_stk[u] = true;
    for (int i = h[u]; i != -1; 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;
            Size[scc_cnt] ++;
        } while (y != u);
    }
}

int main() {
    scanf("%d%d", &n, &m);
    memset(h, -1, sizeof h);
    while (m--) {
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b);
    }
    
    for (int i = 1; i <= n; ++i) {
        if (!dfn[i]) {
            tarjan(i);
        }
    }
    
    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];
            if (a != b) dout[a]++;
        }
    }
    
    int zeros = 0, sum = 0;
    for (int i = 1; i <= scc_cnt; ++i) {
        if (!dout[i]) {
            zeros++;
            sum += Size[i];
            if (zeros > 1) {
                sum = 0;
                break;
            }
        }
    }
    
    printf("%d\n", sum);
    
    return 0;
}

题目2：367学校网络

C++代码如下，

#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 110, M = 10010;

int n;
int h[N], e[M], ne[M], idx;
int dfn[M], low[N], timestamp;
int stk[N], top;
bool in_stk[N];
int id[N], scc_cnt;
int din[N], dout[N];

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, 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;
        }  while (y != u);
    }
}

int main() {
    cin >> n;
    memset(h, -1, sizeof h);
    for (int i = 1; i <= n; ++i) {
        int t;
        while (cin >> t, t) add(i, t);
    }
    
    for (int i = 1; i <= n; ++i) {
        if (!dfn[i]) {
            tarjan(i);
        }
    }
    
    for (int i = 1; i <= n; ++i) {
        for (int j = h[i]; j != -1; j = ne[j]) {
            int k = e[j];
            int a = id[i], b = id[k];
            if (a != b) {
                dout[a]++;
                din[b]++;
            }
        }
    }
    
    int a = 0, b = 0;
    for (int i = 1; i <= scc_cnt; ++i) {
        if (!din[i]) a++;
        if (!dout[i]) b++;
    }
    
    printf("%d\n", a);
    if (scc_cnt == 1) puts("0");
    else printf("%d\n", max(a, b));
    
    return 0;
}

题目3：1175最大半连通子图

C++代码如下，

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_set>

using namespace std;

typedef long long LL;

const int N = 100010, M = 2000010;
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() {
    memset(h, -1, sizeof h);
    memset(hs, -1, sizeof hs);
    
    scanf("%d%d%d", &n, &m, &mod);
    while (m--) {
        int a, b;
        scanf("%d%d", &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;
    }
    
    printf("%d\n", maxf);
    printf("%d\n", sum);
    
    return 0;
}