ZOJ 2182 Cable TV Network（无向图点割-最大流）
题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2182
题意：给出一个无向图，问最少删掉多少个顶点之后图变得不连通？
思路：将原图每个点拆点(i,i+n)，连边<i,i+n,1>，对原图的边(u,v)，连边<u+n,v,INF>,<v+n,u,INF>。然后对于每对顶点(i,j)跑最大流(i+n,j)。所有最大流的最小值即为答案。
struct node
{
int v,cap,next;
};
node edges[N*10];
int head[N],e;
int curedge[N],h[N],num[N],pre[N];
int s,t;
void add(int u,int v,int cap)
{
edges[e].v=v;
edges[e].cap=cap;
edges[e].next=head[u];
head[u]=e++;
}
void Add(int u,int v,int cap)
{
add(u,v,cap);
add(v,u,0);
}
int Maxflow(int s,int t,int n)
{
clr(h,0); clr(num,0);
int i;
FOR0(i,n+1) curedge[i]=head[i];
int u=s,Min,k,x,ans=0;
while(h[u]<n)
{
if(u==t)
{
Min=INF*100;
for(i=s;i!=t;i=edges[curedge[i]].v)
{
x=curedge[i];
if(edges[x].cap<Min)
{
Min=edges[x].cap;
k=i;
}
}
ans+=Min; u=k;
for(i=s;i!=t;i=edges[curedge[i]].v)
{
x=curedge[i];
edges[x].cap-=Min;
edges[x^1].cap+=Min;
}
}
for(i=curedge[u];i!=-1;i=edges[i].next)
{
if(edges[i].cap>0&&h[u]==h[edges[i].v]+1)
{
break;
}
}
if(i!=-1)
{
curedge[u]=i;
pre[edges[i].v]=u;
u=edges[i].v;
}
else
{
if(--num[h[u]]==0) break;
curedge[u]=head[u];
x=n;
for(i=head[u];i!=-1;i=edges[i].next)
{
k=edges[i].v;
if(edges[i].cap>0&&h[k]<x) x=h[k];
}
h[u]=x+1; num[x+1]++;
if(u!=s) u=pre[u];
}
}
return ans;
}
int n,m;
int a[55][55];
int visit[55];
void DFS(int u)
{
visit[u]=1;
int i,v;
FOR1(i,n) if(a[u][i]&&!visit[i])
{
DFS(i);
}
}
int ok()
{
clr(visit,0);
DFS(1);
int i;
FOR1(i,n) if(!visit[i]) return 0;
return 1;
}
int cal(int s,int t)
{
clr(head,-1); e=0;
int i,j;
FOR1(i,n) Add(i,i+n,1);
FOR1(i,n) for(j=1;j<=n;j++) if(a[i][j])
{
Add(i+n,j,INF);
}
return Maxflow(s+n,t,n+n+2);
}
int get()
{
int x=0;
char c=getchar();
while(!isdigit(c))c=getchar();
while(isdigit(c))
{
x=x*10+c-'0';
c=getchar();
}
return x;
}
int main()
{
while(scanf("%d%d",&n,&m)!=-1)
{
if(m==0)
{
if(n==0) puts("0");
else if(n==1) puts("1");
else puts("0");
continue;
}
clr(a,0);
int u,v,i;
FOR0(i,m)
{
u=get(); v=get();
a[u+1][v+1]=a[v+1][u+1]=1;
}
if(!ok())
{
puts("0");
continue;
}
int j;
int ans=INF;
FOR1(i,n) for(j=1;j<=n;j++) if(i!=j)
{
int x=cal(i,j);
ans=min(ans,x);
}
if(ans==INF||ans==n-1) ans=n;
PR(ans);
}
}