poj 1144 Network 联通分量

何为割点？也就是题目中的关键点。在一个无向图中，去掉一个点，这个无向图会变成多个子图，那么这个点就叫做割点

同理，割边也是如此，如果去掉一条边，能让无向图变成多个子图，那么这条边叫做割边，所谓的桥。

 

那么tarjan是如何求的割点的呢？

如果u为割点，当且仅当满足下面的1/2

1、如果u为树根，那么u必须有多于1棵子树

2、如果u不为树根，那么（u,v）为树枝边，当Low[v]>=DFN[u]时。

 

割点的求法倒是看明白了，条件1的意思是若为根，下面如果只有一颗子树，也就是整个图是强连通，那么去掉根节点，肯定不会变成多个子图，因此也不会成为割点。只有大于一颗子树，去掉根节点，才会有两个或者2个以上的子图，从而才能成为割点

 

条件2也比较好理解，u不为树根，那么u肯定有祖先，如果存在Low【v】>=DFN【u】时，表示u的子孙只能通过u才能访问u的祖先，这也就是说，不通过u，u的子孙无法访问u的祖先，那么如果去掉u这个节点，就会至少分为两个子图，一个是u祖先，一个是u子孙的。

Description

A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N . No two places have the same number. The lines are bidirectional and always connect together two places and in each place the lines end in a telephone exchange. There is one telephone exchange in each place. From each place it is 
possible to reach through lines every other place, however it need not be a direct connection, it can go through several exchanges. From time to time the power supply fails at a place and then the exchange does not operate. The officials from TLC realized that in such a case it can happen that besides the fact that the place with the failure is unreachable, this can also cause that some other places cannot connect to each other. In such a case we will say the place (where the failure 
occured) is critical. Now the officials are trying to write a program for finding the number of all such critical places. Help them.

Input

The input file consists of several blocks of lines. Each block describes one network. In the first line of each block there is the number of places N < 100. Each of the next at most N lines contains the number of a place followed by the numbers of some places to which there is a direct line from this place. These at most N lines completely describe the network, i.e., each direct connection of two places in the network is contained at least in one row. All numbers in one line are separated 
by one space. Each block ends with a line containing just 0. The last block has only one line with N = 0;

Output

The output contains for each block except the last in the input file one line containing the number of critical places.

Sample Input

55 1 2 3 4062 1 35 4 6 200

Sample Output

12

Hint

You need to determine the end of one line.In order to make it's easy to determine,there are no extra blank before the end of each line.

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;struct node{    int v,next;} edge[200010];int head[20010],low[20010],dfn[20010],jilu[20010];int cnt,times,n,root;void add(int u,int v){    edge[cnt].v=v;    edge[cnt].next=head[u];    head[u]=cnt++;    edge[cnt].v=u;    edge[cnt].next=head[v];    head[v]=cnt++;}void tarjan(int u,int fa){    int flag=0;    int  son=0;    dfn[u]=low[u]=++times;    for(int e=head[u]; e!=-1; e=edge[e].next)    {        int v=edge[e].v;        if(v==fa&&!flag)        {            flag=1;            continue;        }        if(!dfn[v])        {            tarjan(v,u);            son++;            low[u]=min(low[u],low[v]);           if((u==1&&son>1)||(u!=1&&dfn[u]<=low[v]))                jilu[u]=1;        }        else            low[u]=min(low[u],dfn[v]);    }}int main(){    int ans;    while(~scanf("%d",&n)&&n)    {        int u,v;        memset(head,-1,sizeof(head));        memset(dfn,0,sizeof(dfn));        memset(jilu,0,sizeof(jilu));        times=0;        cnt=0;        while(~scanf("%d",&u)&&u)        {            while(getchar()!='\n')            {                scanf("%d",&v);                add(u,v);            }        }        root=0;        tarjan(1,-1);        ans=0;        for(int i=1; i<=n; i++)        {            if(jilu[i])                ans++;        }        printf("%d\n",ans);    }    return 0;}

下面两个我感觉差不多，但是不知道为什么是错误的，算了开学再研究吧，还是算法不熟悉

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;struct node{    int v,next;} edge[20010];int head[20010],low[20010],dfn[20010],jilu[20010];int cnt,times,n,root;void add(int u,int v){    edge[cnt].v=v;    edge[cnt].next=head[u];    head[u]=cnt++;    edge[cnt].v=u;    edge[cnt].next=head[v];    head[v]=cnt++;}void tarjan(int u,int fa){    int flag=0;    dfn[u]=low[u]=++times;    for(int e=head[u]; e!=-1; e=edge[e].next)    {        int v=edge[e].v;        if(v==fa&&!flag)        {            flag=1;            continue;        }        if(!dfn[v])        {            tarjan(v,u);            low[u]=min(low[u],low[v]);            if(low[v]>dfn[u]&&u!=1)                jilu[u]++;            else if(u==1)            {                root++;            }        }        else            low[u]=min(low[u],dfn[v]);    }}int main(){    int ans;    while(~scanf("%d",&n)&&n)    {        int u,v;        memset(head,-1,sizeof(head));        memset(dfn,0,sizeof(dfn));        memset(jilu,0,sizeof(jilu));        times=0; cnt=0;        while(~scanf("%d",&u)&&u)        {            while(getchar()!='\n')            {                scanf("%d",&v);                add(u,v);            }        }        root=0;        tarjan(1,-1);        ans=0;        if(root>1)        {            ans++;        }        for(int i=2; i<=n; i++)        {            if(jilu[i])                ans++;        }        printf("%d\n",ans);    }    return 0;}