UVA 796 Critical Links(无向图求桥)

题目来源：

UVa Online Judge
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=737

 

求一个连通图中必不可少的路径：

#include<stdio.h>
#include<algorithm>
#include<vector>
#include<string.h>
#define N 1010
using namespace std;
struct node
{
    int x, y;
} no[N];
int dfn[N], low[N];
int f[N], Time, n;
vector<vector<int> >G;
void Init()
{
    G.clear();
    G.resize(n+1);
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(f, 0, sizeof(f));
    Time = 0;
}
void Tarjan(int u, int fa)
{
    int i, len, v;
    dfn[u] = low[u] = ++Time;
    f[u] = fa;
    len = G[u].size();
    for (i = 0; i < len; i++)
    {
        v = G[u][i];
        if (!dfn[v])
        {
            Tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else if (v != fa) low[u] = min(low[u], dfn[v]);
    }
}
int cmp(node a, node b)
{
    if (a.x != b.x) return a.x < b.x;
    return a.y < b.y;
} //从小到大排序
int main ()
{
    int a, b, m, k, i, ni, mn;
    while (scanf("%d", &n) != EOF)
    {
        k = 0;
        Init();
        mn = n;
        while (mn--)
        {
            scanf("%d (%d)", &a, &m);
            while (m--)
            {
                scanf("%d", &b);
                G[a].push_back(b);
                G[b].push_back(a);
            }
        }
        for (i = 0; i < n; i++)
            if (!dfn[i]) Tarjan(i, -1); //可能出现多个连通图的现象，需要每次进行查找，一次查找后该连通图的节点的dfn都不为0，借此可以判断是否还有其他连通图
        for (i = 0; i < n; i++)
        {
            ni = f[i];
            if (ni != -1 && dfn[ni] < low[i]) //父节点为-1说明该点为根节点，与其父节点不可能存在桥
            {                                 //dfn[ni]==low[i]说明i还有其他路径到ni，则说明i-ni不属于桥
                no[k].x = i;
                no[k].y = ni;
                if (i > ni) swap(no[k].x, no[k].y); 
                k++;
            }
        }
        sort(no, no+k, cmp);
        printf("%d critical links\n", k);
        for (i = 0; i < k; i++) printf("%d - %d\n", no[i].x, no[i].y);
        printf("\n");
    }
    return 0;
}