Usaco Section 5.4 telecowmunication (min cut)

Pretty bare a minimal cut. Split each computer into a 1-capacity edge and run the maximum flow. From small to large enumeration of each computer, if removed after the maximum stream = before the maximum stream +1, then this computer is one of the answer.

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#include <cstdio>#include <vector>#include <cstring>#define REP (i,r) for (int i=0;i<r;i++)#define CLR (x,c) memset (x,c,sizeof (x) )#define REP (I,L,R) for (int i=l;i<r;i++)using namespace std;const int maxn=100*2+5;const int inf=1<<30;int x[maxn][2];struct Edge {int from,to,cap,flow;Edge (int u,int v,int c,int f): From (U), to (v), Cap (c), Flow (f) {}};struct ISAP {int n,m,s,t;int P[MAXN];int CUR[MAXN];int NUM[MAXN];int D[MAXN];vector<int> G[MAXN];vector<edge> edges;void init (int n) {this->n=n;Rep (I,n) g[i].clear ();edges.clear ();}int Addedge (int from,int to,int cap) {Edges.push_back (Edge) {from,to,cap,0});Edges.push_back (Edge) {to,from,0,0});m=edges.size ();G[from].push_back (m-2);G[to].push_back (m-1);return m-2;}int augment () {int x=t,a=inf;While (x!=s) {Edge &e=edges[p[x]];a=min (a,e.cap-e.flow);X=edges[p[x]].from;}x=t;While (x!=s) {Edges[p[x]].flow+=a;Edges[p[x]^1].flow-=a;X=edges[p[x]].from;}return A;}int maxflow (int s,int t) {int flow=0;this->s=s; this->t=t;CLR (d,0); CLR (num,0); CLR (cur,0); CLR (p,0) ;Rep (I,edges.size ()) edges[i].flow=0;Rep (i,n) num[d[i]]++;int x=s;While (d[s]<n) {if (x==t) {flow+=augment (); x=s;}int ok=0;Rep (I,cur[x],g[x].size ()) {Edge &e=edges[g[x][i]];if (e.cap>e.flow && d[x]==d[e.to]+1) {ok=1;P[e.to]=g[x][i];cur[x]=i;x=e.to;Break ;}}if (!ok) {int m=n-1;Rep (I,g[x].size ()) {Edge &e=edges[g[x][i]];if (e.cap>e.flow) m=min (m,d[e.to]);}if (--num[d[x]]==0) break;num[d[x]=m+1]++;cur[x]=0;if (x!=s) X=edges[p[x]].from;}}return flow;}} ISAP;int main (){freopen ("telecow.in", "R", stdin);freopen ("Telecow.out", "w", stdout);CLR (x,-1);int n,m,s,t,a,b;scanf ("%d%d%d%d", &n,&m,&s,&t);--s;--t;Isap.init (2*n);Rep (i,n) {X[i][0]=isap.addedge (i,i+n,1);X[i][1]=isap.addedge (i+n,i,1);}Rep (i,m) {scanf ("%d%d", &a,&b);--a;--b;Isap.addedge (a+n,b,m+1);Isap.addedge (b+n,a,m+1);}int Maxflow=isap.maxflow (s+n,t);vector<int> ans; ans.clear ();Rep (i,n) if (i!=s && i!=t) {isap.edges[x[i][0]].cap=0;isap.edges[x[i][1]].cap=0;if (Isap.maxflow (s+n,t) ==maxflow-1) {Ans.push_back (i+1);maxflow--;} else {isap.edges[x[i][0]].cap=1;isap.edges[x[i][1]].cap=1;}if (!maxflow) break;}printf ("%d\n", Ans.size ());printf ("%d", ans[0]);Rep (I,1,ans.size ()) printf ("%d", ans[i]);printf ("\ n");return 0;}

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Telecowmunication

Farmer John ' s cows like-to-keep in touch via email so they has created a network of cowputers so that they can INTERCOWMU Nicate. These machines route email so if there exists a sequence of C Cowputers A1, A2, ..., A (c) such that A1 was connected T o A2, A2 is connected to A3, and so on then A1 and a (c) can send mail to one another.

Unfortunately, a cow would occasionally step on a cowputer or Farmer John would drive over it, and the machine would stop wor King. This means is the cowputer can no longer route email, so connections to and from that Cowputer is no longer usable.

Cows is pondering the minimum number of these accidents that can occur before they can no. longer use their both favori Te cowputers to send e-mail to each of the other. Write a program to calculate the minimal value for them, and the calculate a set of machines that corresponds to the this mini Mum.

For example the network:

               1*               /                

shows 3 cowputers connected with 2 lines. We want to send messages between 1 with 2. Direct lines connect 1-3 and 2-3. If Cowputer 3 is off, them there is no-to-get a message from 1 to 2.Program Name:telecowinput FORMAT

Line 1

Four space-separated integers:n, M, C1, and C2. N is the number of computers (1 <= N <=), which was numbered 1..N. M is the number of connections between pairs of Cowputers (1 <= M <= 600). The last of the numbers, C1 and C2, is the ID numbers of the cowputers that the questioning cows is using. Each connection are unique and bidirectional (if C1 is connected to C2, then C2 are connected to C1). There can is at the most one wire between any and given cowputers. Computer C1 and C2 won't have a direction connection.

Lines 2..m+1

The subsequent M lines contain pairs of Cowputers ID numbers that has connections between them.

SAMPLE INPUT (file telecow.in)

OUTPUT FORMAT

Generate lines of output. The first line was the minimum number of cowputers that can was down before terminals C1 & C2 are no longer connected. The second line was a minimal-length sorted list of cowputers that would cause C1 & C2 to no longer be connected. Note that neither C1 nor C2 can go down. In case of ties, the program should output the set of computers this, if interpreted as a base N number, is the smallest O Ne.

SAMPLE OUTPUT (file telecow.out)

1 3

Usaco Section 5.4 telecowmunication (min cut)