acwing算法提高之图论--无向图的双连通分量

1 介绍

本博客用来记录无向图的双连通分量的相关题目。

以下所有概念都是针对无向图而言的。
桥：本质是边，去掉它，图就不连通了。这样的边叫作桥。
边双连通分量：不包含桥的连通块，且边的数目最大。

割点：本质是结点，去掉它（即去掉这个结点和它所关联的边），图就不连通了。这样的结点叫作割点。
点双连通分量：不包含割点的连通块，且结点的数目最大。

边双连通分量的求解方法：引入时间戳，与有向图强连通分量的求解方法类似。

结论1：对于有向图，至少需要加多少条边，能将此图变成一个强连通分量。答案是max(p, q)，其中p是起点个数（即入度为0的结点个数），q是终点个数（即出度为0的结点个数）。

结论2：对于无向图，至少需要加入多少条边，能将此图变成一个边双连通分量。答案是(cnt + 1) / 2。其中cnt是指出度和入度皆为为1的结点数目。

点双连通分量的求解方法：

2 训练

题目1：395冗余路径

C++代码如下，

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

using namespace std;

const int N = 5010, M = 20010;

int n, m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;
int stk[N], top;
int id[N], dcc_cnt;
bool is_bridge[M];
int d[N];

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

void tarjan(int u, int from) {
    dfn[u] = low[u] = ++timestamp;
    stk[++top] = u;
    
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (!dfn[j]) {
            tarjan(j, i);
            low[u] = min(low[u], low[j]);
            if (dfn[u] < low[j]) {
                is_bridge[i] = is_bridge[i ^ 1] = true;
            }
        } else if (i != (from ^ 1)) {
            low[u] = min(low[u], dfn[j]);
        }
    }
    
    if (dfn[u] == low[u]) {
        ++dcc_cnt;
        int y;
        do {
            y = stk[top--];
            id[y] = dcc_cnt;
        } while (y != u);
    }
}

int main() {
    cin >> n >> m;
    memset(h, -1, sizeof h);
    while (m--) {
        int a, b;
        cin >> a >> b;
        add(a, b), add(b, a);
    }
    
    tarjan(1, -1);
    
    for (int i = 0; i < idx; ++i) {
        if (is_bridge[i]) {
            d[id[e[i]]]++;
        }
    }
    
    int cnt = 0;
    for (int i = 1; i <= dcc_cnt; ++i) {
        if (d[i] == 1) {
            cnt++;
        }
    }
    
    printf("%d\n", (cnt + 1) / 2);
    
    return 0;
}

题目2：1183电力

C++代码如下，

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

using namespace std;

const int N = 10010, M = 30010;

int n, m;
int h[N], e[M], ne[M], idx;
int dfn[M], low[N], timestamp;
int root, ans;

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;
    
    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 (low[j] >= dfn[u]) cnt++;
        } else {
            low[u] = min(low[u], dfn[j]);
        }
    }
    
    if (u != root) cnt++;
    
    ans = max(ans, cnt);
    
    return;
}

int main() {
    while (scanf("%d%d", &n, &m), n || m) {
        memset(dfn, 0, sizeof dfn);
        memset(h, -1, sizeof h);
        idx = timestamp = 0;
        
        while (m--) {
            int a, b;
            scanf("%d%d", &a, &b);
            add(a, b), add(b, a);
        }
        
        ans = 0;
        int cnt = 0;
        for (root = 0; root < n; root++) {
            if (!dfn[root]) {
                cnt++;
                tarjan(root);
            }
        }
        
        printf("%d\n", ans + cnt - 1);
    }
    return 0;
}

题目3：396矿场搭建

C++代码如下，

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

using namespace std;

typedef unsigned long long ULL;

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].push_back(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] = true;
                ++dcc_cnt;
                int y;
                do {
                    y = stk[top--];
                    dcc[dcc_cnt].push_back(y);
                } while (y != j);
                dcc[dcc_cnt].push_back(u);
            }
        } else {
            low[u] = min(low[u], dfn[j]);
        }
    }
}

int main() {
    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(n, a), 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;
        }
        
        printf("Case %d: %d %llu\n", T++, res, num);
    }
    
    return 0;
}