HDU1269 迷宫城堡（tarjan判断强连通）

来源：互联网发布：大数据驱动安全编辑：程序博客网时间：2024/05/17 06:45

裸的判断强连通，如果整个图中只有一个强连通分量，就“Yes”，否则“No”

tarjan的代码：

#include<cstdio>#include<cstring>#include<stack>using namespace std;const int N = 10001;struct Edge{int s,e,next;}edge[10*N];int n,m,e_num,vis_num,cnt,head[N],instack[N],low[N],tim[N];void AddEdge(int a,int b){edge[e_num].s=a; edge[e_num].e=b; edge[e_num].next=head[a]; head[a]=e_num++;}void getmap(){int a,b;memset(instack,0,sizeof(instack));memset(low,0,sizeof(low));memset(tim,-1,sizeof(tim));e_num=0;memset(head,-1,sizeof(head));while(m--){scanf("%d%d",&a,&b);AddEdge(a,b);}}stack <int> st;void tarjan(int x){int j;tim[x]=low[x]=++vis_num;instack[x]=1;st.push(x);for(j=head[x];j!=-1;j=edge[j].next){int u=edge[j].e;if(tim[u]==-1){tarjan(u);if(low[x]>low[u])low[x]=low[u];}else if(instack[u] && low[x]>tim[u])low[x]=tim[u];}if(low[x]==tim[x]){cnt++;do{j=st.top();st.pop();instack[j]=0;}while(j!=x);}}void solve(){int i;vis_num=cnt=0;for(i=1;i<=n;i++){if(tim[i]==-1)tarjan(i);}if(cnt==1)printf("Yes\n");else printf("No\n");}int main(){while(scanf("%d%d",&n,&m),n+m){getmap();solve();}return 0;}