UVA-1660 Cable TV Network (min cut)

The main topic: to a N-point, M-edge of the graph, the minimum point cut set of the cardinality.

Problem Analysis: The base of the minimum point cut set of the graph can be changed to the minimum cut. Considering the graph of single source s single-sink T, if a minimum point set is required, so that the graph is no longer connected after the point set is removed (the number of connected components is increased), simply split each point into two (in and out points), and with a capacity of 1 arc, the other arc unchanged, on the new network to find the smallest cut will get the base of this minimum point set. But the subject is passive without sinks, you can specify a point as the source point, enumerate the other points as sinks, and obtain the n-1 of the point set base, the smallest is the answer. Be aware that each enumeration is re-built.

The code is as follows:

# include<iostream># include<cstdio># include<vector># include<queue># include<cstring ># include<algorithm>using namespace Std;const int inf=1<<30;const int maxn=105;struct Edge{int fr,to,c    AP,FW;        Edge (int fr,int to,int cap,int fw) {this->fr=fr;        this->to=to;        this->cap=cap;    this->fw=fw;    }};struct dinic{vector<edge>edges;    vector<int>g[maxn];    int D[MAXN],CUR[MAXN];    int vis[maxn],n,s,t;        void init (int n) {this->n=n;        Edges.clear ();    for (int i=0;i<n;++i) g[i].clear ();        } void Addedge (int u,int v,int cap) {Edges.push_back (Edge (u,v,cap,0));        Edges.push_back (Edge (v,u,0,0));        int m=edges.size ();        G[u].push_back (m-2);    G[v].push_back (m-1);        } bool BFs () {memset (vis,0,sizeof (VIS));        Vis[s]=1;        d[s]=0;        queue<int>q;        Q.push (s);        while (!q.empty ()) {    int U=q.front ();            Q.pop ();                for (int i=0;i<g[u].size (); ++i) {edge& e=edges[g[u][i]];                    if (E.cap>e.fw&&!vis[e.to]) {vis[e.to]=1;                    d[e.to]=d[u]+1;                Q.push (e.to);    }}} return vis[t]; } int dfs (int u,int a) {if (u==t| |        a==0) return A;        int flow=0,f;            for (int &i=cur[u];i<g[u].size (); ++i) {edge& e=edges[g[u][i]];                if (d[e.to]==d[u]+1&& (F=dfs (E.to,min (A,E.CAP-E.FW))) >0) {e.fw+=f;                Edges[g[u][i]^1].fw-=f;                Flow+=f;                A-=f;            if (a==0) break;    }} return flow;        } int Maxflow (int s,int t) {this->s=s,this->t=t;        int flow=0;            while (BFS ()) {memset (cur,0,sizeof (cur));        Flow+=dfs (S,inf);    } return flow; }};D INIC solver;string p;int n,m,mark[55];vector<edge>e;void addedge (int u,int v,int cap) {E.push_back (Edge (u,v,cap,0)) ;}    void Solve () {int len=p.size (), POS;    for (Pos=0;pos<len;++pos) if (p[pos]== ', ') break;    int a=0,b=0;    for (int i=1;i<pos;++i) a=a*10+p[i]-' 0 ';    for (int i=pos+1;i<len-1;++i) b=b*10+p[i]-' 0 ';    Addedge (A<<1|1,b<<1,inf); Addedge (B<<1|1,a<<1,inf);}        void Look () {for (int i=0;i<solver.edges.size (); ++i) {edge& e=solver.edges[i];    cout<<e.fr<< ' <<e.to<< ' <<e.cap<< ' <<e.fw<<endl;    }}int Main () {int A, B;        while (~SCANF ("%d%d", &n,&m)) {e.clear ();        for (int i=0;i<n;++i) Addedge (i<<1,i<<1|1,1);            while (m--) {cin>>p;        Solve ();        }//look ();        int ans=n,k=e.size ();            for (int i=1;i<n;++i) {solver.init (n*2); for (intJ=0;J&LT;K;++J) {Solver.addedge (E[J].FR,E[J].TO,E[J].CAP);            } int Flow=solver.maxflow (1,i<<1);        Ans=min (Ans,flow);    } printf ("%d\n", ans); } return 0;}

　　

UVA-1660 Cable TV Network (min cut)