BZOJ4435: [Cerc2015]juice Junctions

Maximum flow = minimum cut, and since the degree of the point is not more than 3, the minimum cut does not exceed the complexity of the 3,ek algorithm for $o (N+M) $.

By the division of the minimum cut tree, set $f[i][j][k]$ to indicate the minimum cut to $i$, $j $ point in the $k$ sub-rule is connected with $s$, $h [i][j]$ for $f[i][j][k]$ hash value, then if $h[k][i]=h[k][j]$, It is indicated that the minimum cut between $i$ and $j$ does not exceed $k$.

Time Complexity $o (n (n+m)) $, which requires a large number of constant optimizations.

#include <cstdio> #include <cstring> #define N 3010struct e{short u,v,nxt,f;} E[9010];int n,m,i,j,k,a[n],s,t,ed=1,g[n],h[n],q[n],maxflow,ans;unsigned int hash[4][n],pos=1;inline void Add (int u, int V,int f) {e[++ed].u=u;e[ed].v=v;e[ed].f=f;e[ed].nxt=g[u];g[u]=ed;}  inline bool BFs () {int*l=q,*r=q+1,i;  memset (h+1,-1,sizeof (int) *n);  h[q[0]=s]=0;  while (L<r) for (I=G[*L++];I;I=E[I].NXT) if (H[E[I].V]&LT;0&AMP;&AMP;E[I].F) h[*r++=e[i].v]=i; return h[t]!=-1;}  void Solve (int l,int r) {if (l>=r) return;  int i,j;  for (i=2;i<=ed;i++) e[i].f=e[i^1].f=1;  s=a[r],t=a[l],maxflow=0;  while (BFS ()) for (maxflow++,i=t;i!=s;i=e[j].u) e[j=h[i]].f--, e[j^1].f++;  unsigned int*p=hash[maxflow]+1;  for (pos*=233,i=1;i<=n;p++) if (~h[i++]) *p+=pos;  Int*l=q+l,*r=q+r,*k=a+l;  while (K<=a+r) if (h[*k]<0) *l++=*k++;else *r--=*k++;  memcpy (a+l,q+l,sizeof (int) * (r-l+1)); Solve (L,R-Q), Solve (l-q,r);}  int main () {scanf ("%d%d", &n,&m);  while (m--) scanf ("%d%d", &i,&j), add (i,j,1), add (j,i,1); for (I=1; i<=n;i++) a[i]=i;  Solve (1,n);  for (i=1;i<=n;i++) for (j=i+1;j<=n;j++) for (k=0;k<4;k++) if (Hash[k][i]!=hash[k][j]) {ans+=k;break;} Return printf ("%d", ans), 0;}

　　

BZOJ4435: [Cerc2015]juice Junctions