POJ 1523 双联通求割点

                                                                                              SPF

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5025		Accepted: 2311

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

无向图割点模板题：

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
using namespace std;
int head[2000],tol,low[2000],dfn[2000],Stack[3000],instack[2000],add_bloak[2000],indexx,top,cut[2000],n;
struct node
{
        int to,next;
}edge[100000];
map<int,int> mp,mm;
void add(int u,int v)
{
        edge[tol].to=v;
        edge[tol].next=head[u];
        head[u]=tol++;
}
void tarjin(int u,int pre)
{
        int i,v;
        low[u]=dfn[u]=++indexx;
        Stack[top++]=u;
        instack[u]=1;
        int son=0;
        for(i=head[u];i!=-1;i=edge[i].next)
        {
                v=edge[i].to;
                if(v==pre)continue;
                if(!dfn[v])
                {
                        son++;
                       tarjin(v,u);
                       if(low[u]>low[v])low[u]=low[v];
                       if(u!=pre&&low[v]>=dfn[u])
                       {
                               cut[u]=1;
                               add_bloak[u]++;
                       }
                }
                else if(low[u]>dfn[v])low[u]=dfn[v];
        }
        if(u==pre)add_bloak[u]=son-1;
        if(u==pre&&son>1)cut[u]=1;
        instack[u]=0;
        top--;
}
void solve()
{
        memset(dfn,0,sizeof(dfn));
        memset(instack,0,sizeof(instack));
        memset(add_bloak,0,sizeof(add_bloak));
        memset(cut,0,sizeof(cut));
        indexx=top=0;
        tarjin(1,1);
        int sum=0;
        for(int i=1;i<=n;i++)if(cut[i])sum++;
        if(sum==0){puts("  No SPF nodes\n");return;}
        for(int i=1;i<=n;i++)
        {
                if(!cut[i])continue;
                printf("  SPF node %d leaves %d subnets\n",mm[i],add_bloak[i]+1);
        }
        puts("");
}
int find(int x)
{
        if(mp[x])return mp[x];
        mp[x]=++n;
        mm[n]=x;
        return mp[x];
}
int main()
{
        int i,j,k,m,t=0,p,q;
        while(~scanf("%d",&p)&&p)
        {
                ++t;scanf("%d",&q);
                printf("Network #%d\n",t);
                n=0;mp.clear();mm.clear();
                memset(head,-1,sizeof(head));tol=0;
                p=find(p);q=find(q);
                //cout<<p<<" "<<q<<endl;
                add(p,q);
                add(q,p);
                while(~scanf("%d",&p)&&p)
                {
                        scanf("%d",&q);
                        p=find(p);
                        q=find(q);
                        //cout<<p<<" "<<q<<endl;
                        add(p,q);
                        add(q,p);
                }
               // cout<<tol<<endl;
               solve();
        }
        return 0;
}