UVA 796 Critical Links

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http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82833#problem/C

UVA链接地址

Root

796 - Critical Links

Time limit: 3.000 seconds

题目大意：给你一个网络要求这里面的桥。

输入数据：

n 个点

点的编号（与这个点相连的点的个数m）依次是m个点的

输入到文件结束。

桥输出的时候需要排序

知识汇总：

桥：无向连通图中，如果删除某条边后，图变成不连通了，则该边为桥。

求桥：

在求割点的基础上吗，假如一个边没有重边（重边 1-2, 1->2 有两次，那么 1->2 就是有两条边了，那么 1->2就不算是桥了）。

当且仅当 (u,v) 为父子边，且满足 dfn[u] < low[v]

#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-8struct 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]);    }}

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