Summary of global minimum cut learning

The global minimum cut means to take s and T in an all-in-one graph, and to find the minimum value of minimum cut.

There is also a description of the minimum cost of deleting edges in undirected graphs so that they do not connect

Of course, this topic can be solved by splitting + minimum cut

But the complexity of time is around O (n^4).

A better solution can be solved in the time complexity of O (n^3)

The practice is this:

First, for any two points in the graph s->t

Either S and T are not the answer in a set, the answer is clearly the smallest cut of s and T

Otherwise s and T in a set, we can shrink s and t into a point, it is not difficult to prove that this is equivalent

We simulate this process, each time the S and T run the minimum cut, the time complexity is probably the same as the division + minimum cut almost Owo

But we notice that both S and T are either, that is, we just need to ask for a set of solutions.

This gives you a construction algorithm:

1, at the beginning a set is empty, we take a little to join a set of

2, the definition W (a,x) represents a set of all points to x Benquan and

3, each time looking for W (a,x) The largest point (the same word whichever), add a set and update the other W value

4, the last two points added are S and T, the minimum cut is the last addition of W (A,T)

This process directly simulates an O (n^2), plus a pinch point for up to O (n) times

So the total time complexity O (n^3) can be optimized with heaps, but dense graphs run very slowly and slowly

Bzoj 3345

 #include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include < Algorithm> using namespace std; const int Maxn=510;const int Oo=0x7fffffff;int n,m,u,v,d;int s,t,cut;int f[maxn][maxn];int w[maxn];bool com[maxn],vis[ MAXN]; void Get_ans () {cut=oo;    S=t=-1;    memset (vis,false,sizeof (VIS));    Memset (W,0,sizeof (w));        for (int i=1;i<=n;++i) {int mx=-oo,tmp;            for (int j=1;j<=n;++j) {if (!com[j]&&!vis[j]&&w[j]>mx) {mx=w[j];tmp=j;        }} if (t==tmp) return; s=t;        t=tmp;        Cut=mx;vis[tmp]=true;    for (int j=1;j<=n;++j) if (!com[j]&&!vis[j]) w[j]+=f[tmp][j]; }return;}    int get_sw () {memset (com,false,sizeof (COM));    int Ans=oo;        for (int i=1;i<n;++i) {Get_ans ();        Ans=min (Ans,cut);        if (ans==0) return 0;        Com[t]=true;        for (int j=1;j<=n;++j) {if (!com[j]) {f[s][j]+=f[t][j];f[j][s]+=f[j][t];}  }  }return ans;}    int main () {scanf ("%d%d", &n,&m);    memset (F,0,sizeof (f));        for (int i=1;i<=m;++i) {scanf ("%d%d%d", &u,&v,&d);    F[u][v]+=d;f[v][u]+=d;    } printf ("%d\n", GET_SW ()); return 0;}

HDU 3691

Notice here that S is given by the

But it's not hard to see that there are no eggs, the answer is the global minimum cut

Because the final global minimum cut must have a T and s not belong to a set Owo

And because it is a global minimum cut, the minimum cut of this set of s->t is the global minimum cut

#include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include < Algorithm>using namespace Std;typedef long long ll;const int maxn=310;const LL oo=1ll<<60;int n,m,k;int u,v,w;in T s,t; LL Ans,cut; LL F[MAXN][MAXN];    LL w[maxn];bool com[maxn],vis[maxn];void Read (int &num) {Num=0;char Ch=getchar (); while (ch< '! ')    Ch=getchar (); while (ch>= ' 0 ' &&ch<= ' 9 ') num=num*10+ch-' 0 ', Ch=getchar ();} void Get_ans () {cut=oo;    S=t=-1;    memset (vis,false,sizeof (VIS));    memset (w,0,sizeof (W));        for (int i=1;i<=n;++i) {LL mx=-oo;int tmp;        for (int j=1;j<=n;++j) {if (!com[j]&&!vis[j]&&w[j]>mx) {mx=w[j];tmp=j;}        } if (t==tmp) return; s=t;        t=tmp;        Cut=mx;vis[tmp]=true;        for (int j=1;j<=n;++j) {if (!com[j]&&!vis[j]) {w[j]+=f[tmp][j];} }}return;}    LL get_sw () {ans=oo;memset (com,0,sizeof (COM)); for (int i=1;i<n;++i) {Get_ans ();       Ans=min (Ans,cut);        if (ans==0) return ans;        Com[t]=true;        for (int j=1;j<=n;++j) {if (!com[j]) {f[s][j]+=f[t][j];f[j][s]+=f[j][t];} }}return ans;        int main () {while (scanf ("%d%d%d", &n,&m,&k) ==3) {if (!n&&!m&&!k) break;        memset (F,0,sizeof (f));            for (int i=1;i<=m;++i) {read (u); Read (v); Read (w);        F[u][v]+=w;f[v][u]+=w;    } printf ("%d\n", GET_SW ()); }return 0;}

　　

Summary of global minimum cut learning