POJ1523(求连用分量数目,tarjan算法原理理解)

#include"cstdio"
#include"cstring"
using namespace std;
const int MAXN=1005;
struct Edge{
    int to,next;
}es[MAXN*2];
int V,E;
int head[MAXN];
inline int max(int u,int v)
{
    return u > v? u: v;
}
inline int min(int a,int b)
{
    return a > b? b: a;
}
void add_edge(int u,int v)
{
    es[E].to=v;
    es[E].next=head[u];
    head[u]=E++;
    V=max(max(u,v),V);
}
bool flag;
int root;
int subnets[MAXN];
int index;
int dfn[MAXN],low[MAXN];
void tarjan(int u,int fa)
{
    int son=0;
    dfn[u]=low[u]=++index;
    for(int i=head[u];i!=-1;i=es[i].next)
    {
        int v=es[i].to;
        if(!dfn[v])
        {
            tarjan(v,u);
            son++;
            low[u]=min(low[u],low[v]);
            if((u==root&&son>1)||(u!=root&&dfn[u]<=low[v]))
            {
                flag=true;
                subnets[u]++;
                //u->v  该边导致u成为割点 
                //当dfn[u]==low[v]时u->v为返祖边,u、v处于同一双连通分量中 
                //当dfn[u]<low[v]时u->v为割边
                //删除割点u产生的连通数目为:u所在的连通分量数目+与u所连接的割边的数目+1(边:fa->u) 
            }
        }
        else if(v!=fa) low[u]=min(low[u],dfn[v]);
    }
}
int main()
{
    int cas=0;
    int u,v;
    while(true)
    {
        v=-1;
        index=0;
        memset(subnets,0,sizeof(subnets));
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        memset(head,-1,sizeof(head));
        V=-1,E=0;
        flag=false;
        while(scanf("%d",&u)&&u)
        {
            scanf("%d",&v);
            add_edge(u,v);
            add_edge(v,u);
        }
        if(v==-1)    break;
        root=V;
        tarjan(root,-1);
        printf("Network #%d\n",++cas);
        if(flag)
        {
            for(int i=1;i<=V;i++)
            {
                if(subnets[i]>0)
                {
                    printf("  SPF node %d leaves %d subnets\n",i,subnets[i]+1);//加上fa->u该边所连接的连通分量 
                }
            }
        }
        else    printf("  No SPF nodes\n");
        printf("\n");
    }
    return 0;
}