The story of zjoi2009 Wolf and goat

Since this question is so watery, I will not write it ......

The essence of fence Mining: it can only be built in two adjacent places, and after the establishment, there is no path between the wolf and the goat. Cut is defined to make the s set and t set do not have a path. The problem requires the least barrier to be built, so it is the problem of least cutting.

From the source point to all the wolves connected to an ∞ edge, from all the sheep to the sink point connected to an ∞ edge, this can ensure that the wolf and the sheep are in different vertex sets. Then, from the wolf to the adjacent sheep and the open space, the open space to the adjacent open space and the sheep are connected to a side with a traffic of 1, and the maximum flow can be minimized.

Or link all vertices to four sides .. It's a long time -- hzwer

Code: (from hzwer)

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #define inf 0x7fffffff 5 #define T 10001 6 using namespace std; 7 int head[10005],q[10005],h[10005]; 8 int cnt=1,ans,n,m; 9 int xx[4]={0,0,1,-1},yy[4]={1,-1,0,0},mp[105][105];10 struct data{int to,next,v;}e[500001];11 void ins(int u,int v,int w)12 {e[++cnt].to=v;e[cnt].next=head[u];e[cnt].v=w;head[u]=cnt;}13 void insert(int u,int v,int w)14 {ins(u,v,w);ins(v,u,0);}15 bool bfs()16 {17      int t=0,w=1,i,now;18      memset(h,-1,sizeof(h));19      q[0]=0;h[0]=0;20      while(t<w)21      {22                now=q[t];t++;i=head[now];23                while(i)24                {25                        if(e[i].v&&h[e[i].to]==-1)26                        {27                                                  h[e[i].to]=h[now]+1;28                                                  q[w++]=e[i].to;29                                                  }30                        i=e[i].next;31                        }32                }33     return h[T]==-1? 0:1;34  }35 int dfs(int x,int f)36 {37     if(x==T)return f;38     int w,used=0,i;39     i=head[x];40     while(i)41     {42             if(e[i].v&&h[e[i].to]==h[x]+1)43             {44                                           w=f-used;45                                           w=dfs(e[i].to,min(w,e[i].v));46                                           e[i].v-=w;47                                           e[i^1].v+=w;48                                           used+=w;49                                           if(used==f)return f;50                                           }51             i=e[i].next;52             }53     if(!used)h[x]=-1;54     return used;55 }56 void dinic(){while(bfs())ans+=dfs(0,inf);}57 void ini()58 {59      scanf("%d%d",&n,&m);60      for(int i=1;i<=n;i++)61          for(int j=1;j<=m;j++)62              scanf("%d",&mp[i][j]);63      }64 void build()65 {66      for(int i=1;i<=n;i++)67          for(int j=1;j<=m;j++)68          {69              if(mp[i][j]==1)insert(0,(i-1)*m+j,inf);70              else if(mp[i][j]==2)insert((i-1)*m+j,T,inf);71              for(int k=0;k<4;k++)72              {73                      int nowx=i+xx[k],nowy=j+yy[k];74                      if(nowx<1||nowx>n||nowy<1||nowy>m||mp[i][j]==2)continue;75                      if(mp[i][j]!=1||mp[nowx][nowy]!=1)76                      insert((i-1)*m+j,(nowx-1)*m+nowy,1);77                      }78              }79      }80 int main()81 {82     ini();83     build();84     dinic();85     printf("%d",ans);86     return 0;87     }

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