uva796 Critical Links（求桥并按边输出）

最新推荐文章于 2017-01-12 19:41:41 发布

围巾的ACM

最新推荐文章于 2017-01-12 19:41:41 发布

阅读量438

点赞数

CC 4.0 BY-SA版权

分类专栏：图论-有无向图连通问题

本文链接：https://blog.youkuaiyun.com/qq_21057881/article/details/50639265

图论-有无向图连通问题专栏收录该内容

30 篇文章

订阅专栏

本文介绍了一种用于寻找无向图中所有桥的算法，并通过具体的C++代码实现展示了如何求解。该算法采用Tarjan算法的变体，通过深度优先搜索（DFS）来识别连接点之间的关键边。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

思路：模板题...求桥并按边输出

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <queue>
#include <stack>
#include <vector>
#include <map>

using namespace std;

#define N 12005
#define INF 0xfffffff
#define PI acos (-1.0)
#define EPS 1e-8

struct node
{
    int x, y;
    bool friend operator < (node a, node b)
    {
        if (a.x == a.y) return a.y < b.y;
        return a.x < b.x;
    }
}bridge[N];

vector <int> G[N];
int n, low[N], dfn[N], f[N], Time, ans;

void Init ();
void solve ();
void tarjan (int u, int fa);

int main ()
{
    while (~scanf ("%d", &n))
    {
        Init ();
        for (int i=0; i<n; i++)
        {
            int a, b, m;
            scanf ("%d (%d)", &a, &m);
            while (m--)
            {
                scanf ("%d", &b);
                G[a].push_back (b);
                G[b].push_back (a);
            }
        }
        solve ();
    }
    return 0;
}

void Init ()
{
    memset (low, 0, sizeof (low));
    memset (dfn, 0, sizeof (dfn));
    memset (f, 0, sizeof (f));
    Time = 0;
    for (int i=0; i<n; i++)
        G[i].clear ();
}

void solve ()
{
    ans = 0;
    for (int i=0; i<n; i++)
        if (!low[i])
            tarjan (i, -1);

    for (int i=0; i<n; i++)
    {
        int v = f[i];
        if (v != -1 && dfn[v] < low[i])
        {
            bridge[ans].x = i;
            bridge[ans].y = v;
            if (bridge[ans].x > bridge[ans].y)
                swap (bridge[ans].x, bridge[ans].y);
            ans++;
        }
    }

    sort (bridge, bridge+ans);

    printf ("%d critical links\n", ans);
    for (int i=0; i<ans; i++)
        printf ("%d - %d\n", bridge[i].x, bridge[i].y);
    puts ("");
}

void tarjan (int u, int fa)
{
    low[u] = dfn[u] = ++Time;
    f[u] = fa;
    int len = G[u].size (), v;

    for (int i=0; i<len; i++)
    {
        v = G[u][i];
        if (!low[v])
        {
            tarjan (v, u);
            low[u] = min (low[u], low[v]);
        }
        else if (fa != v)
            low[u] = min (low[u], dfn[v]);
    }
}