Describle 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 5 5 1 2 3 4 0 6 2 1 3 5 4 6 2 0 0 1 2 3 4 5 6 7 8 Sample Output 1 2  

題目大意：給出一個n點無向圖，求圖中割點個數

可以用Tarjan演算法求割點

對於根節點，判斷是不是割點，只需要計算其子樹數量，如果有2棵即以上的子樹，就是割點。因為如果去掉這個點，這兩棵子樹就不能互相到達。

對於非根節點，對於邊(u, v)，如果low[v]>=dfn[u]，此時u就是割點。

用鄰接矩陣去重邊

程式碼：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
using namespace std;
const int N = 100+20;
struct node
{
    int v,next;
} e[N * N];
bool instack[N],cut[N];
int first[N],dfn[N],low[N];
int g[N][N];
int cnt, tot, time, root;
void adde (int u, int v)
{
    e[tot].v = v;
    e[tot].next = first[u];
    first[u] = tot++;
}
void tarjan (int u, int pre)
{
    dfn[u] = low[u] = ++time;
    instack[u] = 1;
    int cnt = 0;
    for (int i = first[u]; ~i; i = e[i].next)
    {
        int v = e[i].v;
        if (v == pre) continue;
        if (!dfn[v])
        {
            tarjan(v, u);
            cnt++;
            if (low[u] > low[v]) low[u] = low[v];
            if (root == u && cnt > 1) cut[u] = 1; //判斷根節點是否是割點 
            else if (u != root && low[v] >= dfn[u]) cut[u] = 1; //判斷非根節點是否是割點
        }
        else 
            low[u] = min(low[u], dfn[v]);
    }
}
void init ()
{
    memset (dfn, 0, sizeof(dfn));
    memset (low, 0, sizeof(low));
    memset (instack, 0, sizeof(instack));
    memset (cut, 0, sizeof(cut));
    memset (first, -1, sizeof(first));
    memset(g,0,sizeof(g));
    tot =time = 0;
}
int main()
{
    int n;
    int u, v;
    while (~scanf("%d", &n), n)
    {
        init();
        while (scanf("%d", &u)&&u)
        {
            while (getchar() != '\n')
            {
                scanf("%d", &v);
                g[u][v]=1;
                g[v][u]=1;
            }
        }
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
            if(g[i][j])
            adde(i,j);
        root = 1;
        tarjan (1, -1);
        int ans = 0;
        for (int i = 1; i <= n; ++i)
            if (cut[i])
                ans++;
        printf("%d\n", ans);
    }
    return 0;
}