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这道题是求哪个点是割点,并且能把原图分成几块,add_block数组是求增加的所以还需要加1,然后就是我一直在PE,后来发现需要每输出一组后,有一个空行
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
const int MAXN=1005;
const int MAXM=0x3f3f3f;
struct edge
{
int to,next;
bool cut;
} edge[MAXM];
int head[MAXN],tot;
int low[MAXN],dfn[MAXN],Stack[MAXN];
int Index,top;
bool Instack[MAXN];
bool cut[MAXN];
int add_block[MAXN];
int bridge;
void addedge(int u,int v)
{
edge[tot].to=v;
edge[tot].next=head[u];
edge[tot].cut=false;
head[u]=tot++;
}
void tarjan(int u,int pre)
{
int v;
low[u]=dfn[u]=++Index;
Stack[top++]=u;
Instack[u]=true;
int son=0;
for(int i=head[u]; i!=-1; i=edge[i].next)
{
v=edge[i].to;
if(v==pre)
continue;
if(!dfn[v])
{
son++;
tarjan(v,u);
if(low[u]>low[v])
low[u]=low[v];
if(u!=pre&&low[v]>=dfn[u])
{
cut[u]=true;
add_block[u]++;
}
}
else if(low[u]>dfn[v])
low[u]=dfn[v];
}
if(u==pre&&son>1)
cut[u]=true;
if(u==pre)
add_block[u]=son-1;
Instack[u]=false;
top--;
}
void init()
{
memset(head,-1,sizeof(head));
memset(cut,0,sizeof(cut));
memset(Instack,0,sizeof(Instack));
memset(add_block,0,sizeof(add_block));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
}
int a,b,n;
int main()
{
int cases=1;
while(cin>>a)
{
init();
if(a!=0)
{
cin>>b;
addedge(a,b);
addedge(b,a);
}
else if(a==0)
break;
while(1)
{
cin>>a;
if(a!=0)
{
cin>>b;
addedge(a,b);
addedge(b,a);
}
else if(a==0)
break;
}
tarjan(1,1);
bool flag=0;
cout<<"Network #"<<cases++<<endl;
for(int i=1; i<=1005; i++)
{
if(cut[i])
{
cout<<" SPF node "<<i<<" leaves "<<add_block[i]+1<<" subnets"<<endl;
flag=1;
}
}
if(flag==0)
cout<<" No SPF nodes"<<endl;
cout<<endl;
}
return 0;
}
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原文地址:http://blog.csdn.net/zcmartin2014214283/article/details/51331686