This is the minimum cut, but I think for a long time ....
Consider all sheep connected to the source point, the wolf connected to the meeting point, adjacent points on both sides of the minimum cut (maximum flow) can be.
Correctness? Considering something like a dichotomy, we just need to set the wolf together and waist the sheep.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define Maxe 100005
#define MAXV 10011
#define INF 1234567
#define S 0
#define T 10010
using namespace Std;
int n,m,map[105][105],dx[]={0,0,1,0,-1},dy[]={0,1,0,-1,0};
int dis[maxv],g[maxv],nume=1;
struct Edge
{
int v,f,nxt;
}e[maxe];
void Addedge (int u,int v,int flow)
{
E[++nume].v=v;
E[nume].nxt=g[u];
E[nume].f=flow;
G[u]=nume;
E[++nume].v=u;
E[NUME].NXT=G[V];
e[nume].f=0;
G[v]=nume;
}
void Build ()
{
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
{
if (map[i][j]==1) Addedge (S, (i-1) *m+j,inf);
else if (map[i][j]==2) Addedge ((i-1) *m+j,t,inf);
for (int k=1;k<=4;k++)
{
int tx=i+dx[k],ty=j+dy[k];
if ((tx>=1) && (tx<=n) && (ty>=1) && (ty<=m))
Addedge ((i-1) *m+j, (tx-1) *m+ty,1);
}
}
}
BOOL BFs ()
{
memset (dis,-1,sizeof (dis));
Queue <int> q;
Q.push (s);
dis[s]=0;
while (!q.empty ())
{
int Head=q.front ();
Q.pop ();
for (int i=g[head];i;i=e[i].nxt)
{
if ((e[i].f>0) && (dis[e[i].v]<0))
{
dis[e[i].v]=dis[head]+1;
Q.push (E[I].V);
}
}
}
if (dis[t]==-1) return false;
return true;
}
int dinic (int x,int low)
{
if (x==t) return low;
int ret=0;
for (int i=g[x];low && I;I=E[I].NXT)
{
if ((e[i].f!=0) && (dis[e[i].v]==dis[x]+1))
{
int Dd=dinic (E[i].v,min (LOW,E[I].F));
LOW=LOW-DD;
RET=RET+DD;
E[I].F=E[I].F-DD;
E[I^1].F=E[I^1].F+DD;
}
}
if (ret==0) dis[x]=-1;
return ret;
}
int main ()
{
memset (map,0,sizeof (map));
memset (G,0,sizeof (g));
scanf ("%d%d", &n,&m);
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
scanf ("%d", &map[i][j]);
Build ();
int ans=0;
while (BFS () ==true)
Ans=ans+dinic (S,inf);
printf ("%d\n", ans);
return 0;
}
Bzoj 1412 The story of the Wolf and the Sheep