3 3 1 2 2 3 3 1 3 3 1 2 2 3 3 2 0 0
Yes No
#include <iostream> #include<cstdio> #include<cstring> #include<set> using namespace std; set<int> s[100005],sv; int vis[100005]; int n,m,i,x,y,flag; void dfs(int ori,int room) { set<int>::iterator it=s[room].begin(); while(it!=s[room].end()) { if (vis[*it]) { it++; continue;} vis[*it]=1; sv.insert(*it); dfs(ori,*it); it++; } return; } int main() { while(~scanf("%d%d",&n,&m),n) { for(i=1;i<=n;i++) s[i].clear(); for(int i=1;i<=m;i++) { scanf("%d%d",&x,&y); s[x].insert(y); } flag=1; for(i=1;i<=n;i++) { memset(vis,0,sizeof(vis)); sv.clear(); vis[i]=1; dfs(i,i); if(sv.size()!=n-1) {flag=0; break;} } if (flag) printf("Yes\n"); else printf("No\n"); } return 0; }
tarjan算法。。
#include <iostream> #include<cstring> #include<cstdio> #include<deque> #include<algorithm> using namespace std; int dfn[10005],low[10005],instack[10005],vis[10005],p[10005]; struct node { int num,next; }tree[100005]; int i,j,n,m,x,y,index,l,flag,num; deque<int> s; void tarjan(int u) { dfn[u]=low[u]=++index; vis[u]=1; instack[u]=1; s.push_back(u); for(int i=p[u];i!=-1;i=tree[i].next) { if (!vis[tree[i].num]) { tarjan(tree[i].num); if (flag) return; low[u]=min(low[u],low[tree[i].num]); } else if (instack[tree[i].num]) { low[u]=min(low[u],dfn[tree[i].num]); } } if (dfn[u]==low[u]) { num=0; int v; do { v=s.back(); s.pop_back(); instack[v]=0; num++; } while(u!=v); flag=1; return; } return; } int main() { while(scanf("%d%d",&n,&m) && n!=0) { l=0; memset(p,-1,sizeof(p)); for(i=1;i<=m;i++) { scanf("%d%d",&x,&y); tree[++l].num=y; tree[l].next=p[x]; p[x]=l; } memset(vis,0,sizeof(vis)); memset(instack,0,sizeof(instack)); index=0; flag=0; s.clear(); tarjan(x); if (n==num) printf("Yes\n"); else printf("No\n"); } return 0; }
原文:http://www.cnblogs.com/stepping/p/5648396.html