割点算法
example
1 2
1 3
1 4
2 5
2 6
3 7
3 8
4 9
4 8
0
下面是算法运行过程
from 1 enter to the adj ares 2
from 2 enter to the adj ares 5
end 5
return 2 is a cut node ,find subnet 1
from 2 enter to the adj ares 6
end 6
return 2 is a cut node ,find subnet 2
end 2
from 1 enter to the adj ares 3
from 3 enter to the adj ares 7
end 7
return 3 is a cut node ,find subnet 1
from 3 enter to the adj ares 8
from 8 enter to the adj ares 4
from 4 enter to the adj ares 9
end 9
return 4 is a cut node ,find subnet 1
end 4
end 8
end 3
return 1 is a cut node ,find subnet 1
end 1
运行结果：
Network #1
SPF node 1 leaves 2 subnets
SPF node 2 leaves 3 subnets
SPF node 3 leaves 2 subnets
SPF node 4 leaves 2 subnets
最后subnet数目要加一，因为cut[k]是孤立subnet，还要加上大后方
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
int cut[1001],c[1001],ancestor[1001],d[1001],part[1001];
vector <int> adj[1001];
int ans;
int MAX,n,m;
void input(int n)
{
if(n==0)return;
scanf("%d",&m);
adj[n].push_back(m);
adj[m].push_back(n);
MAX=max(MAX,max(m,n));
}
void DFS(int k,int fa,int deep)
{
int tot;
c[k]=-1;
d[k]=deep;
ancestor[k]=deep,tot=0;
for(int i=0;i<adj[k].size();i++)
{
int w=adj[k][i];
if(w!=fa&&c[w]==-1)ancestor[k]=min( ancestor[k], d[w] );
else if(c[w]==0)
{
//cout<<"from "<<k<<" enter to the adj ares "<<w<<endl;
DFS(w,k,deep+1);
tot++,ancestor[k]=min(ancestor[k],ancestor[w]);
if(fa==-1&&tot>1||fa!=-1&&ancestor[w]>=d[k])
{
cut[k]++;//用来标记是否为割点及能分成几块连通区域(+1)
//cout<<"return "<<k<<" is a cut node ,find subnet "<<cut[k]<<endl;
}
}
}
c[k]=1;
//cout<<"end "<<k<<endl;
}
int main()
{
int cas=0;
while(scanf("%d",&n)==1)
{
if(n==0)break;
for(int i=1;i<1001;i++)
adj[i].clear();
MAX=-1;;
input(n);
while(scanf("%d",&n))
{
if(n==0)break;
input(n);
}
memset(d,0,sizeof(d));
memset(c,0,sizeof(c));
memset(cut,0,sizeof(cut));
memset(ancestor,0,sizeof(ancestor));
DFS(1,-1,0);
if(cas++)printf("\n");
int ex=0;
printf("Network #%d\n",cas);
for(int i=1;i<=MAX;i++)
{
if(cut[i]){printf(" SPF node %d leaves %d subnets\n",i,cut[i]+1);ex=1;}
}
if(!ex)printf(" No SPF nodes\n");
}
}