点双连通分量的求解 Tarjan算法的拓展

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问题描述：

给出一张连通的无向图输出图中的所有连通分量

代码：

#include<iostream>#include<cstdio>#include<cstring>#define maxn 1050using namespace std;struct node{    int from,to,next,vis;    int equall(node b)    {        if((from==b.from)&&to==b.to||(from==b.to&&to==b.from))            return 1;        return 0;    }}edge[maxn*100],Edge;     //int head[maxn];           //邻接表 int e_num;               //int dfn[maxn],low[maxn];int num;node stk[maxn*100];int top;int ss;void init(){    memset(head,-1,sizeof(head));    memset(dfn,0,sizeof(dfn));    num=1;    e_num=0;    top=-1;    ss=0;}void addedge(int a,int b){    edge[e_num]={a,b,head[a],0};    head[a]=e_num++;    edge[e_num]={b,a,head[b],0};    head[b]=e_num++;}void Tarjan(int u){    dfn[u]=low[u]=num++;    for(int i=head[u];i!=-1;i=edge[i].next)    {        if(!edge[i].vis)        {            edge[i].vis=edge[i^1].vis=1;        //防止回边(同一条边)重复入栈            stk[++top]=edge[i];                 //压栈            int v=edge[i].to;            if(!dfn[v])            {                Tarjan(v);                low[u]=min(low[u],low[v]);                if(dfn[u]<=low[v])               //发现u是割点 -----构成一个连通分量                {                    printf("连通分量%d:\n",++ss);                    while(1)                    {                        if(top<0)                            break;                        printf("%d-%d ",stk[top].from,stk[top].to);                        top--;                        if(edge[i].equall(stk[top+1]))                            break;                    }                    cout<<endl;                }            }            else low[u]=min(low[u],dfn[v]);        }    }}int main(){    int n,m;    int a,b;    while(1)    {        printf("请输入点数和边数:\n");        scanf("%d%d",&n,&m);        init();        for(int i=1;i<=m;i++)        {            printf("请输入第%d条边:\n",i);            scanf("%d%d",&a,&b);            addedge(a,b);        }        Tarjan(1);    }    return 0;}

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