POJ3352边双连通分量缩点添加最少无向边构造边双

Posted 2021-08-25 hesorchen

题目

POJ 3352

给一个无向连通图，问至少添加几条边使得去掉图中任意一条边不改变图的连通性（即使得它变为边双连通图）。

求解

1. 对原图求出边双联通分量。
2. 边双缩点成树。
3. 答案为 $\frac{叶子节点数量+1}{2}$。(依次把两个最远的叶子连边)

代码

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
using namespace std;
int n, m;

const int N = 1010;
vector<int> edge[N];
map<int, map<int, int>> mp;
int fa[N], dfn[N], low[N], degree[N], ct;
void tarjan(int u)
{
    dfn[u] = low[u] = ++ct;
    for (int i = 0; i < edge[u].size(); i++)
    {
        int v = edge[u][i];
        if (!dfn[v])
        {
            fa[v] = u;
            tarjan(v);
            low[u] = min(low[u], low[v]);
            if (low[v] > low[u])
                mp[u][v] = mp[v][u] = 1; //u-v 是桥
        }
        else if (v != fa[u])
        {
            low[u] = min(low[u], dfn[v]);
        }
    }
}
int cnt;
int belong[N];
void dfs(int u)
{
    belong[u] = cnt;
    for (int i = 0; i < edge[u].size(); i++)
    {
        int v = edge[u][i];
        if (mp[u][v])
            continue;
        if (dfn[v])
        {
            dfn[v] = 0;
            dfs(v);
        }
    }
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= m; i++)
    {
        int u, v;
        cin >> u >> v;
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    tarjan(1); //求桥
    for (int i = 1; i <= n; i++)
    {
        if (dfn[i])
        {
            ++cnt;
            dfn[i] = 0;
            dfs(i); //边双缩点
        }
    }

    if (cnt == 1) //已经是边双
    {
        puts("0");
    }
    else
    {
        mp.clear();
        for (int i = 1; i <= n; i++)
        {
            for (int j = 0; j < edge[i].size(); j++)
            {
                int v = edge[i][j];
                if (mp[belong[v]][belong[i]] || belong[i] == belong[v])
                    continue;
                mp[belong[v]][belong[i]] = mp[belong[i]][belong[v]] = 1;
                degree[belong[v]]++;
                degree[belong[i]]++;
            }
        }
        int sum = 0;
        for (int i = 1; i <= n; i++)
        {
            sum += degree[belong[i]] == 1; //度为1的节点为叶节点
            degree[belong[i]] = -1;
        }
        cout << (sum + 1) / 2 << endl;
    }
    return 0;
}