传送门

分析

起点向狼连边，羊向终点连边，边权均为inf

每个点向它四联通的点连边权萎1的边

跑最小割即可

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
const int dx[] = {1,-1,0,0};
const int dy[] = {0,0,1,-1};
const int inf = 1e9+7;
int n,m,s,t,head[2000100],to[2000100],nxt[2000100],w[2000100],ano[2000100],cnt,level[2000100],cur[2000100];
inline void add(int x,int y,int z){
    nxt[++cnt]=head[x];
    head[x]=cnt;
    to[cnt]=y;
    w[cnt]=z;
    ano[cnt]=cnt+1;
    nxt[++cnt]=head[y];
    head[y]=cnt;
    to[cnt]=x;
    w[cnt]=0;
    ano[cnt]=cnt-1;
}
inline int id(int x,int y){return (x-1)*m+y;}
inline void go(int x,int y){
    for(int i=0;i<4;i++){
      if(dx[i]+x<=0||dx[i]+x>n||dy[i]+y<=0||dy[i]+y>m)continue;
      add(id(x,y),id(dx[i]+x,dy[i]+y),1);
    }
}
inline bool bfs(){
    memset(level,-1,sizeof(level));
    queue<int>q;
    level[s]=0;
    q.push(s);
    while(!q.empty()){
      int x=q.front();
      q.pop();
      for(int i=head[x];i;i=nxt[i])
        if(level[to[i]]==-1&&w[i]){
          level[to[i]]=level[x]+1;
          if(to[i]==t)return 1;
          q.push(to[i]);
        }
    }
    return 0;
}
inline int dfs(int x,int flow){
    if(x==t||!flow)return flow;
    int res=0;
    cur[x]=head[x];
    for(int i=cur[x];i;i=nxt[i]){
      cur[x]=i;
      if(level[to[i]]==level[x]+1&&w[i]){
          int f=dfs(to[i],min(w[i],flow-res));
          w[i]-=f;
          res+=f;
          w[ano[i]]+=f;
      }
    }
    if(!res)level[x]=-1;
    return res;
}
int main(){
    int i,j,k,Ans=0;
    scanf("%d%d",&n,&m);
    s=n*m+1,t=s+1;
    for(i=1;i<=n;i++)
      for(j=1;j<=m;j++){
        scanf("%d",&k);
        if(k==1)add(s,id(i,j),inf);
          else if(k==2)add(id(i,j),t,inf);
        go(i,j);
      }
    while(bfs())while(int a=dfs(s,inf))Ans+=a;
    cout<<Ans;
    return 0;
}