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);
    }
}

ZOJ 2182 Cable TV Network（无向图点割-最大流）,布布扣,bubuko.com