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
Figure 1: Critical links 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
割桥其实与割点大体思路是一样
割桥只要连接的下一个点不会返祖 那么这个边就是桥
而割点则是只要连接的所有点中存在一个不返祖的边那么该点就为割点
值得注意的是割点需要特判根节点 若根节点存在两个及以上的子节点 那么根节点就是割点
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int first[1005];
int nxt[2000005];
int to[2000005];
int cutline[2000005];//桥biaozhi 1biaoshi 能回去
int dfn[1005];
int low[1005];
int belong[2000005];
int id;
struct node
{
int st;
int en;
} a[2000005];
void tarjan(int x,int last)
{
dfn[x]=low[x]=++id;
for(int i=first[x]; i!=-1; i=nxt[i])
{
int tmp=to[i];
if(!dfn[tmp])//连接的子节点 能回去 不是桥 反之 是桥
{
tarjan(tmp,x);
low[x]=min(low[x],low[tmp]);
if(low[tmp]>dfn[x])
{
cutline[i]=1;
cutline[i^1]=1;
//int cao=i^1;
//cout<<i<<" "<< cao <<endl;
}
}
else if(tmp!=last)
{
low[x]=min(low[x],dfn[tmp]);
}
}
return ;
}
int cmp(node p,node q)
{
if(p.st==q.st)
{
p.en<q.en;
}
return p.st<q.st;
}
int main()
{
int n;
int cnt=0;
while(~scanf("%d",&n))
{
id=0;
memset(to,-1,sizeof(to));
memset(belong,-1,sizeof(belong));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(first,-1,sizeof(first));
memset(nxt,-1,sizeof(nxt));
memset(cutline,0,sizeof(cutline));
cnt=-1;
int temp;
int temp1,temp2;
for(int i=1; i<=n; i++)
{
scanf("%d",&temp1);
getchar();
getchar();
scanf("%d",&temp);
getchar();
while(temp--)
{
scanf("%d",&temp2);
if(temp1>temp2) continue;
cnt++;
belong[cnt]=temp1;
nxt[cnt]=first[temp1];
first[temp1]=cnt;
to[cnt]=temp2;
cnt++;
belong[cnt]=temp2;
nxt[cnt]=first[temp2];
first[temp2]=cnt;
to[cnt]=temp1;
//cout<<cnt<<endl;
}
}
int ans=0;
for(int i=0; i<n; i++)
{
if(!dfn[i]) tarjan(i,i);
}
for(int i=0; i<=cnt; i++)
{
if(cutline[i]==1&&belong[i]<to[i])
{
ans++;
a[ans].st=belong[i];
a[ans].en=to[i];
}
}
sort(a+1,a+1+ans,cmp);
printf("%d critical links
",ans);
for(int i=1; i<=ans; i++)
{
printf("%d - %d
",a[i].st,a[i].en);
}
printf("
");
}
}