标签:com poj can root roo txt ace void www
http://poj.org/problem?id=1144
题意:给你一些点,某些点直接有边,并且是无向边,求有多少个点是割点
割点:就是在图中,去掉一个点,无向图会构成多个子图,这就是割点
Tarjan算法求割点的办法
Low[v]>=dnf[u]也就是说明U的子孙点只能通过U点访问U的祖先点
1 #include <stdio.h> 2 #include <stack> 3 #include <string.h> 4 #define maxn 505 5 6 using namespace std; 7 8 stack <int >s; 9 10 int head[maxn],n,pos,dfn[maxn],low[maxn],bcnt,dindex,num[maxn],root; 11 12 bool vis[maxn]; 13 14 struct node{ 15 int next,to; 16 }edge[maxn]; 17 18 void add(int u,int v) 19 { 20 edge[pos].to = v; 21 edge[pos].next = head[u]; 22 head[u] = pos++; 23 } 24 25 void Tarjan(int u) 26 { 27 dfn[u] = low[u] = ++dindex; 28 vis[u] = true; 29 s.push(u); 30 for(int i = head[u]; i != -1 ; i = edge[i].next) 31 { 32 int v = edge[i].to; 33 if(!vis[v]) 34 { 35 Tarjan(v); 36 if(low[v]<low[u]) low[u] = low[v]; 37 if(low[v]>=dfn[u]&&u!=1) 38 { 39 num[u]++; 40 }else if(u==1) 41 root++; 42 }else if(dfn[v]<low[u]) 43 low[u] = dfn[v]; 44 } 45 } 46 47 int main() 48 { 49 int u,v,ans; 50 // freopen("in.txt","r",stdin); 51 while(scanf("%d",&n),n) 52 { 53 54 memset(head,-1,sizeof(head)); 55 memset(vis,false,sizeof(vis)); 56 memset(dfn,0,sizeof(dfn)); 57 memset(low,0,sizeof(low)); 58 memset(num,0,sizeof(num)); 59 pos = 1; 60 ans = 0; 61 while(scanf("%d",&u)&&u) 62 { 63 while(getchar()!=‘\n‘) 64 { 65 scanf("%d",&v); 66 add(u,v); 67 add(v,u); 68 } 69 } 70 bcnt = dindex = root=0; 71 for(int i = 1;i<=n;i++) 72 if(!dfn[i]) Tarjan(i); 73 for(int i = 1 ; i<=n;i++) 74 if(num[i]) ans++; 75 if(root>1) ans++; 76 printf("%d\n",ans); 77 } 78 return 0; 79 }
https://www.byvoid.com/blog/scc-tarjan/一个很好的学习Tarjan的博客
标签:com poj can root roo txt ace void www
原文地址:http://www.cnblogs.com/Tree-dream/p/6051765.html