POJ1523（tarjan割点后，连通分量有多少）

SPF

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 9693 Accepted: 4365

Description

Consider the two networks shown below. Assuming that data moves around these networks only between directly connected nodes on a peer-to-peer basis, a failure of a single node, 3, in the network on the left would prevent some of the still available nodes from
communicating with each other. Nodes 1 and 2 could still communicate with each other as could nodes 4 and 5, but communication between any other pairs of nodes would no longer be possible. 

Node 3 is therefore a Single Point of Failure (SPF) for this network. Strictly, an SPF will be defined as any node that, if unavailable, would prevent at least one pair of available nodes from being able to communicate on what was previously a fully connected
network. Note that the network on the right has no such node; there is no SPF in the network. At least two machines must fail before there are any pairs of available nodes which cannot communicate. 

Input

The input will contain the description of several networks. A network description will consist of pairs of integers, one pair per line, that identify connected nodes. Ordering of the pairs is irrelevant; 1 2 and 2 1 specify the same connection. All node numbers
will range from 1 to 1000. A line containing a single zero ends the list of connected nodes. An empty network description flags the end of the input. Blank lines in the input file should be ignored.

Output

For each network in the input, you will output its number in the file, followed by a list of any SPF nodes that exist. 

The first network in the file should be identified as "Network #1", the second as "Network #2", etc. For each SPF node, output a line, formatted as shown in the examples below, that identifies the node and the number of fully connected subnets that remain when
that node fails. If the network has no SPF nodes, simply output the text "No SPF nodes" instead of a list of SPF nodes.

Sample Input

1 2

5 4

3 1

3 2

3 4

3 5

0

1 2

2 3

3 4

4 5

5 1

0

1 2

2 3

3 4

4 6

6 3

2 5

5 1

0

0

Sample Output

Network #1

  SPF node 3 leaves 2 subnets

Network #2

  No SPF nodes

Network #3

  SPF node 2 leaves 2 subnets
  SPF node 3 leaves 2 subnets

题意就是求下割点，求过割点后，删除一个割点后两桶分量有几个。求个点我在以前博客里写过了，代码上关于割点后得到连通分量的注释

//去掉一个割点后，根节点特殊，他的所有子树都是一个连通分量，而其他割点
//大家想其他节点是怎么求的，通过看dfn[v]是不是小于等于low[u]，如果小于等于就代表v必须经过u，那么u把图分成了两部分
//以v未根节点的树和u上边的祖先的那一部分，那么我们求得dfn[v]<=low[u]以后就aa[u]++，求u下边的那一部分，最后加1（祖先那一部分）就是割点后的连通分量
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
int to;
int next;
} e[100005];
int r,cnt,sum;
int first[10005],dfn[10005],low[100005],aa[100005];
void init()
{
memset(first,-1,sizeof(first));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(aa,0,sizeof(aa));
r=0;
cnt=1;
sum=0;
}
void add(int u,int v)
{
e[r].to=v;
e[r].next=first[u];
first[u]=r++;
}
void tarjan(int u,int from)
{
dfn[u]=low[u]=cnt++;
for(int i=first[u]; i!=-1; i=e[i].next)
{
int v=e[i].to;
if(dfn[v]==0)
{
tarjan(v,u);
low[u]=min(low[u],low[v]);
if(u==1)
{
sum++;
}
else if(low[v]>=dfn[u])//得到除了根节点以外的割点
{
aa[u]++;
}
}
else if(v!=from)
low[u]=min(low[u],dfn[v]);
}
}
int main()
{
int t=1;
while(1)
{
init();
int u;
int maxx=-1;
scanf("%d",&u);
if(u==0)
break;
int v;
scanf("%d",&v);
if(u>maxx) maxx=u;
if(v>maxx) maxx=v;
add(u,v);
add(v,u);
while(1)
{
scanf("%d",&u);
if(u==0) break;
scanf("%d",&v);
if(u>maxx) maxx=u;
if(v>maxx) maxx=v;
add(u,v);
add(v,u);
}
if(t>1)
printf("\n");
printf("Network #%d\n", t++);
tarjan(1,-1);
if(sum>1) aa[1]=sum-1;//特判根节点为割点情况
int flag=0;
for(int i=1; i<=maxx; i++)
{
if(aa[i])
{
flag=1;
printf(" SPF node %d leaves %d subnets\n",i,aa[i]+1);
}
}
if(!flag) printf(" No SPF
a2f0
nodes\n");
}
}