poj 1144 割点

最基本的求割点的题目，建议深入理解该dfs的过程，还有就是对于同一个点来说，满足它是割点的条件可能会成立多次，所以一定要在dfs结束后统计割点个数，而不是在dfs的过程中实现。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 200;
const int M = 500;
int dfn[N];
int low[N];
int head[N];
bool is_cut[N];
int n, e, dfs_clock;

void init()
{
    e = dfs_clock = 0;
    memset( head, -1, sizeof(head) );
    memset( dfn, 0, sizeof(dfn) );
    memset( is_cut, 0, sizeof(is_cut) );
}

struct Edge 
{
    int v, next;
} edge[M];

void addEdge( int u, int v )
{
    edge[e].v = v;
    edge[e].next = head[u];
    head[u] = e++;
}

void dfs( int u, int fa )
{
    low[u] = dfn[u] = ++dfs_clock;
    int cnt = 0;
    for ( int i = head[u]; i != -1; i = edge[i].next )
    {
        int v = edge[i].v;
        if ( !dfn[v] )
        {
            cnt++;
            dfs( v, u );
            if ( low[v] >= dfn[u] ) is_cut[u] = true;
            low[u] = min( low[u], low[v] );
        }
        else if ( v != fa )
        {
            low[u] = min( low[u], dfn[v] );
        }
    }
    if ( fa < 0 && cnt == 1 ) is_cut[u] = false;
}

int main ()
{
    while ( scanf("%d", &n), n )
    {
        init();
        int u, v;
        while ( scanf("%d", &u), u )
        {
            while ( getchar() != '\n' )
            {
                scanf("%d", &v);
                addEdge( u, v );
                addEdge( v, u );
            }
        }
        dfs( 1, -1 );
        int ans = 0;
        for ( int i = 1; i <= n; i++ )
        {
            if ( is_cut[i] ) ans++;
        }
        printf("%d\n", ans);
    }
    return 0;
}