uva315 Network 【图论-tarjan-求桥】

Description

In a computer network a link L, which interconnects two servers, is considered critical if there are at least two servers A and B such that all network interconnection paths between A and B pass through L. Removing a critical link generates two disjoint sub–networks such that any two servers of a sub–network are interconnected. For example, the network shown in figure 1 has three critical links that are marked bold: 0 -1, 3 - 4 and 6 - 7.



It is known that:

1. the connection links are bi–directional;

2. a server is not directly connected to itself;

3. two servers are interconnected if they are directly connected or if they are interconnected with the same server;

4. the network can have stand–alone sub–networks.

Write a program that finds all critical links of a given computer network.

Input

The program reads sets of data from a text file. Each data set specifies the structure of a network and

has the format:

no of servers

server0 (no of direct connections) connected server … connected server…

serverno of servers (no of direct connections) connected server … connected server

The first line contains a positive integer no of servers(possibly 0) which is the number of network

servers. The next no of servers lines, one for each server in the network, are randomly ordered and

show the way servers are connected. The line corresponding to serverk, 0 ≤ k ≤ no of servers − 1,

specifies the number of direct connections of serverk and the servers which are directly connected to

serverk. Servers are represented by integers from 0 to no of servers − 1. Input data are correct. The

first data set from sample input below corresponds to the network in figure 1, while the second data

set specifies an empty network.

Output

The result of the program is on standard output. For each data set the program prints the number of

critical links and the critical links, one link per line, starting from the beginning of the line, as shown

in the sample output below. The links are listed in ascending order according to their first element.

The output for the data set is followed by an empty line.

Sample Input

8

0 (1) 1

1 (3) 2 0 3

2 (2) 1 3

3 (3) 1 2 4

4 (1) 3

7 (1) 6

6 (1) 7

5 (0)

0

Sample Output

3 critical links

0 - 1

3 - 4

6 - 7

0 critical links

题目大意：给一个无向图，求桥的个数，并按照顺序输出各个桥



桥的定义：无向连通图的桥是指，删除该边(i,j)后顶点i和j不再连通。上图中的桥有：(1,3)，(1,2)，(5,6)，(9,7)。

桥的性质和求法：如果y是x的儿子且Ancestory>Dx（注意是严格大于），那么删除边(x,y)后，顶点y将与x不连通。

AC代码：

# include <stdio.h>
# include <string.h>
# include <vector>
# include <algorithm>

using namespace std;

# define MAXN 100005

struct Edge
{
int to;
int next;
int cut;
}edge[MAXN];

int DFN[MAXN];
int LOW[MAXN];
int head[MAXN], tot;
int Index;
int bridge;

void Init()
{
tot = 0;
memset(head, -1, sizeof(head));

}

void addedge(int u, int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
edge[tot].cut = 0;
head[u] = tot++;
}

void tarjan(int u, int pre)
{
int v;
LOW[u] = DFN[u] = ++Index;
for (int i = head[u]; i != -1; i = edge[i].next)
{
v = edge[i].to;
if (v == pre)
{
continue;
}
if (!DFN[v])
{
tarjan(v, u);
if (LOW[u] > LOW[v])
{
LOW[u] = LOW[v];
}

//一条无向边(u,v)是桥，当且仅当(u,v)为树枝边，且满足DFS(u)<Low(v)。
if (DFN[u] < LOW[v])
{
bridge++;
edge[i].cut = 1;
edge[i^1].cut = 1;
}
}
else if (LOW[u] > DFN[v])
{
LOW[u] = DFN[v];
}
}
}

void solve(int n)
{
int i;
Index = 0;
bridge = 0;
memset(DFN, 0, sizeof(DFN));
memset(LOW, 0, sizeof(LOW));
for (i = 1; i <= n; i++)
{
if (!DFN[i])
{
tarjan(i, i);
}
}
printf("%d critical links\n", bridge);
vector <pair<int, int> > ans;
for (int u = 1; u <= n; u++)
{
for (i = head[u]; i != -1; i = edge[i].next)
{
if (edge[i].cut && edge[i].to > u)
{
ans.push_back(make_pair(u, edge[i].to));
}
}
}
sort(ans.begin(), ans.end());
for (i = 0; i < ans.size(); i++)
{
printf("%d - %d\n", ans[i].first-1, ans[i].second-1);
}
printf("\n");
}

int main(void)
{
int n;
while (~scanf("%d", &n))
{
int u, k, v;
int i;
Init();
for (i = 0; i < n; i++)
{
scanf("%d (%d)", &u, &k);
u++;
while (k--)
{
scanf("%d", &v);
v++;
if (v <= u)
{
continue;
}
addedge(u, v);
addedge(v, u);
}
}
solve(n);
}
return 0;
}