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★★ 输入文件:gd.in
输出文件:gd.out
简单对比
时间限制:1 s 内存限制:128 MB
输入文件名: gd.in
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输出文件名:gd.out
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1 #include<iostream> 2 using namespace std; 3 #include<cstdio> 4 #include<cstring> 5 #define N 102 6 int low[N],dfn[N],root,topt=0,n,a,b,ans[N]; 7 bool is_gd[N]; 8 struct Edge{ 9 int v,last; 10 }edge[N*N*2]; 11 int head[N],t=0; 12 void add_edge(int u,int v) 13 { 14 ++t; 15 edge[t].v=v; 16 edge[t].last=head[u]; 17 head[u]=t; 18 } 19 void input() 20 { 21 scanf("%d",&n); 22 while(scanf("%d%d",&a,&b)==2) 23 { 24 add_edge(a,b); 25 add_edge(b,a); 26 } 27 } 28 void tarjan(int k) 29 {/*记住这个模板就好了*/ 30 int cnt=0; 31 dfn[k]=low[k]=++topt; 32 for(int l=head[k];l;l=edge[l].last) 33 { 34 int v=edge[l].v; 35 if(!dfn[v]) 36 { 37 tarjan(v); 38 ++cnt; 39 low[k]=min(low[k],low[v]); 40 if((root==k&&cnt>1)||(root!=k&&low[v]>=dfn[k])) 41 is_gd[k]=true; 42 } 43 else low[k]=min(low[k],dfn[v]); 44 } 45 } 46 int main() 47 { 48 freopen("gd.in","r",stdin); 49 freopen("gd.out","w",stdout); 50 input(); 51 root=1; 52 tarjan(1); 53 for(int i=1;i<=n;++i) 54 if(is_gd[i]) 55 ans[++ans[0]]=i; 56 printf("%d\n",ans[0]); 57 for(int i=1;i<=ans[0];++i) 58 printf("%d\n",ans[i]); 59 fclose(stdin);fclose(stdout); 60 return 0; 61 }
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原文地址:http://www.cnblogs.com/c1299401227/p/5564996.html