http://poj.org/problem?id=3180
还牛举行正式舞会,无语 题意写的那么纠结干嘛
就是求一个有向图中 所含节点个数 大于等于2的强连通分量的个数
代码:
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#define Min(a,b)a<b?a:b
#define nMAX 10002
#define mMAX 50002
using namespace std;
int head[nMAX],dfn[nMAX],low[nMAX],stack[nMAX],node_sta[nMAX];//栈中点的个数
int s_edge,n,top,atype,times;
bool insta[nMAX];
struct
{
int to,next;
}edge[mMAX];
void addedge(int u,int v)
{
s_edge++;
edge[s_edge].to=v;
edge[s_edge].next=head[u];
head[u]=s_edge;
return ;
}
void tarjan(int u)
{
dfn[u]=++times;
low[u]=times;
insta[u]=1;
stack[++top]=u;
int e;
for(e=head[u];e;e=edge[e].next)
{
int v=edge[e].to;
if(!dfn[v])
{
tarjan(v);
low[u]=Min(low[u],low[v]);
}
else if(insta[v])//!!!
low[u]=Min(low[u],dfn[v]);
}
int j;
if(low[u]==dfn[u])
{
atype++;
do
{
j=stack[top--];
insta[u]=0;
node_sta[atype]++;
}while(j!=u);
}
return ;
}
void init()
{
s_edge=0;
top=0;
atype=0;
times=0;
memset(head,0,sizeof(head));
memset(insta,0,sizeof(insta));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(node_sta,0,sizeof(node_sta));
return ;
}
int main()
{
int m,u,v,i;
while(~scanf("%d%d",&n,&m))
{
init();
while(m--)
{
scanf("%d%d",&u,&v);
addedge(u,v);
}
for(i=1;i<=n;i++)
if(!dfn[i])
tarjan(i);
int ans=0;
for(i=1;i<=atype;i++)
if(node_sta[i]>=2)
ans++;
printf("%d\n",ans);
}
return 0 ;
}