Uva 796 Critical Links (割边+排序)

题目链接：

　　Uva 796 Critical Links

题目描述：

　　题目中给出一个有可能不连通的无向图，求出这个图的桥，并且把桥按照起点升序输出（还有啊，还有啊，每个桥的起点要比终点靠前啊），这个题目读了好几遍，但是依旧没有找到数据范围写在哪里，经过无数次runtime error最终把范围定在1W左右。（题目晦涩难懂，wa到死了，嗷~~~~~~）

解题思路：

　　就是用Tarjan算法dfs出low和dfn数组，然后记录下来起点和终点排好序的桥，然后把桥排序输出就ok了。

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int maxn = ;
 struct node
 {
     int to, next;
 } edge[maxn*];
 bool cmp (node a, node b)
 {
     if (a.next == b.next)
         return a.to < b.to;
     return a.next < b.next;
 }
 int low[maxn], dfn[maxn], head[maxn];
 int Father[maxn], tot, ntime, cnt;
 node Q[maxn * ];
 void init ()
 {
     ntime = tot = cnt = ;
     memset (low, , sizeof(low));
     memset (dfn, , sizeof(dfn));
     memset (head, -, sizeof(head));
     memset (Father, , sizeof(Father));
 }
 void Add (int from, int to)
 {
     edge[tot].to = to;
     edge[tot].next = head[from];
     head[from] = tot ++;
 }
 void Tarjan (int u, int father)
 {
     low[u] = dfn[u] = ++ntime;
     Father[u] = father;
     for (int i=head[u]; i!=-; i=edge[i].next)
     {
         int v = edge[i].to;
         if (!dfn[v])
         {
             Tarjan (v, u);
             low[u] = min (low[u], low[v]);
         }
        else if (father != v)
             low[u] = min (low[u], dfn[v]);
     }
 }
 int main ()
 {
     int n, m;
     node q;
     while (scanf ("%d", &n) != EOF)
     {
         init ();
         m = n;
         while (n --)
         {
             int u , k;
             scanf ("%d (%d)", &u, &k);
             while (k --)
             {
                 int v;
                 scanf ("%d", &v);
                 Add (u, v);
                 Add (v, u);
             }
         }
         for (int i=; i<m; i++)
             if (!dfn[i])
                 Tarjan (i, -);

         for (int i=; i<m; i++)
         {
             int v = Father[i];
             if (v != - && dfn[v] < low[i])
             {
                 q.next = v;
                 q.to = i;
                 if (q.next > q.to)
                     swap(q.next, q.to);
                 Q[cnt++] = q;
             }
         }
         printf ("%d critical links\n", cnt);
         sort (Q, Q+cnt, cmp);
         for (int i=; i<cnt; i++)
             printf ("%d - %d\n", Q[i].next, Q[i].to);
         printf ("\n");
     }
     return ;
 }