这里先给出结论:
对于非根节点的割点,它能分割图为cut[i]+1个连通分量;对于根节点割点,它能分割图为cut[i]个连通分量
至于为什么,找了好久没找到答案...
F - SPF
题意:
求删点割点后,割点所割连通分量数
#include <cstdio>
#include<cstring>
#include<vector>
#include<stack>
#include<set>
#include<algorithm>
#include<iterator>
using namespace std;
const int MAXN=1010;
const int MAXE=200010;
struct Node
{
int to,next;
}edge[MAXE];
int head[MAXN],cnt;
void addEdge(int u,int v)
{
edge[cnt].to=v;
edge[cnt].next=head[u];
head[u]=cnt++;
}
int cutCnt[MAXN],low[MAXN],dfn[MAXN],clocks;
set<int> cut;
void DFS(int u,int e)
{
low[u]=dfn[u]=++clocks;
int child=0;
for(int i=head[u];i!=-1;i=edge[i].next)
{
if(i==(e^1)) continue;
int v=edge[i].to;
if(dfn[v]==0)
{
child++;
DFS(v,i);
low[u]=min(low[v],low[u]);
if(low[v]>=dfn[u])
{
cut.insert(u);
cutCnt[u]++;
}
}
else if(dfn[v]<dfn[u])
{
low[u]=min(low[u],dfn[v]);
}
}
if(e<0&&child==1)
{
set<int>::iterator it=cut.find(u);
if(it!=cut.end()) cut.erase(it);
}
else if(e!=-1) cutCnt[u]++;//非根节点
}
void work(int n)
{
memset(dfn,0,sizeof(dfn));
memset(cutCnt,0,sizeof(cutCnt));
clocks=0;
cut.clear();
for(int i=1;i<=n;i++)
{
if(dfn[i]==0)
{
DFS(i,-1);
}
}
}
int main()
{
int n,m,a,b,cas=1,maxv;
while(true)
{
maxv=0;
scanf("%d",&a);
maxv=max(maxv,a);
if(a==0) break;
scanf("%d",&b);
maxv=max(maxv,b);
memset(head,-1,sizeof(head));
cnt=0;
addEdge(a,b);
addEdge(b,a);
while(true)
{
scanf("%d",&a);
maxv=max(maxv,a);
if(a==0) break;
scanf("%d",&b);
maxv=max(maxv,b);
addEdge(a,b);
addEdge(b,a);
}
work(maxv);
printf("Network #%d\n",cas++);
if(cut.size()==0) printf(" No SPF nodes\n");
else
{
for(set<int>::iterator it=cut.begin();it!=cut.end();it++)
{
printf(" SPF node %d leaves %d subnets\n",*it,cutCnt[*it]);
}
}
printf("\n");
}
}
其实一开始不知道有上面那个结论,所以我最初的做法是先求出所有的割点和点连通分量,然后再对每个割点求从该割点出发的每条边的另一个端点的不同的点连通分量数。
#include <cstdio>
#include<cstring>
#include<vector>
#include<stack>
#include<set>
#include<algorithm>
#include<iterator>
using namespace std;
const int MAXN=1010;
const int MAXE=200010;
struct Node
{
int to,next;
}edge[MAXE];
int head[MAXN],cnt;
void addEdge(int u,int v)
{
edge[cnt].to=v;
edge[cnt].next=head[u];
head[u]=cnt++;
}
struct Edge
{
int u,v;
Edge(){}
Edge(int u,int v):u(u),v(v){}
};
int cutCnt[MAXN],low[MAXN],dfn[MAXN],clocks,bccno[MAXN],bcc_cnt;
int vis[MAXN];
stack<Edge> sta;
set<int> cut;
void DFS(int u,int e)
{
low[u]=dfn[u]=++clocks;
int child=0;
for(int i=head[u];i!=-1;i=edge[i].next)
{
if(i==(e^1)) continue;
int v=edge[i].to;
Edge ee(u,v);
if(dfn[v]==0)
{
sta.push(ee);
child++;
DFS(v,i);
low[u]=min(low[v],low[u]);
if(low[v]>=dfn[u])
{
cut.insert(u);
bcc_cnt++;
while(true)
{
Edge x=sta.top();
sta.pop();
if(bccno[x.u]!=bcc_cnt) {bccno[x.u]=bcc_cnt;}
if(bccno[x.v]!=bcc_cnt) {bccno[x.v]=bcc_cnt;}
if(x.u==u&&x.v==v) break;
}
}
}
else if(dfn[v]<dfn[u])
{
sta.push(ee);
low[u]=min(low[u],dfn[v]);
}
}
if(e<0&&child==1)
{
set<int>::iterator it=cut.find(u);
if(it!=cut.end()) cut.erase(it);
}
}
void work(int n)
{
memset(dfn,0,sizeof(dfn));
clocks=bcc_cnt=0;
memset(bccno,0,sizeof(bccno));
while(!sta.empty()) sta.pop();
cut.clear();
for(int i=1;i<=n;i++)
{
if(dfn[i]==0)
{
DFS(i,-1);
}
}
int sum;
for(set<int>::iterator it=cut.begin();it!=cut.end();it++)
{
int u=*it;
sum=0;
memset(vis,0,sizeof(vis));
for(int j=head[u];j!=-1;j=edge[j].next)
{
int v=edge[j].to;
if(vis[bccno[v]]==0)
{
vis[bccno[v]]=1;
sum++;
}
}
cutCnt[u]=sum;
}
}
int main()
{
int n,m,a,b,cas=1,maxv;
while(true)
{
maxv=0;
scanf("%d",&a);
maxv=max(maxv,a);
if(a==0) break;
scanf("%d",&b);
maxv=max(maxv,b);
memset(head,-1,sizeof(head));
cnt=0;
addEdge(a,b);
addEdge(b,a);
while(true)
{
scanf("%d",&a);
maxv=max(maxv,a);
if(a==0) break;
scanf("%d",&b);
maxv=max(maxv,b);
addEdge(a,b);
addEdge(b,a);
}
work(maxv);
printf("Network #%d\n",cas++);
if(cut.size()==0) printf(" No SPF nodes\n");
else
{
for(set<int>::iterator it=cut.begin();it!=cut.end();it++)
{
printf(" SPF node %d leaves %d subnets\n",*it,cutCnt[*it]);
}
}
printf("\n");
}
}
