\(Solution\)
考虑\(1\)和\(2\)相邻是必须要长度为\(1\)的篱笆,所以只需考虑\(1,2\)和\(0\)之间要不要放篱笆,用\(S\)连向所有的\(1\),所有的\(2\)连向\(T\),现在问题是要使\(S\)与\(T\)不连通,这不是经典的最小割问题吗?
\(Code\)
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf = 2147483647;
int h[10005],cur[10005],n,m,dep[10005],q[10005],tot = 1,S,T,ans,a[105][105];
struct edge{
int to,nxt,z;
}e[2000005];
void add(int x,int y,int z)
{
e[++tot] = edge{y,h[x],z},h[x] = tot;
e[++tot] = edge{x,h[y],0},h[y] = tot;
}
int bfs()
{
for (int i = 1; i <= T; i++) cur[i] = h[i];
memset(dep,0,sizeof dep);
int head = 0,tail = 1;
q[1] = S,dep[S] = 1;
while (head < tail)
{
int u = q[++head];
for (int i = h[u]; i; i = e[i].nxt)
{
int v = e[i].to;
if (dep[v] || e[i].z <= 0) continue;
dep[v] = dep[u] + 1,q[++tail] = v;
}
}
return dep[T];
}
int dinic(int u,int mn,int fa)
{
if (u == T || mn <= 0) return mn;
int flow = 0,sum = 0;
for (int &i = cur[u]; i; i = e[i].nxt)
{
int v = e[i].to;
if (dep[v] != dep[u] + 1 || v == fa || e[i].z <= 0) continue;
sum = dinic(v,min(mn,e[i].z),u);
if (sum <= 0) continue;
flow += sum,mn -= sum,e[i].z -= sum,e[i ^ 1].z += sum;
if (mn <= 0) break;
}
return flow;
}
void work(int x,int y,int x1,int y1)
{
int k1 = (x - 1) * m + y,k2 = (x1 - 1) * m + y1;
if (a[x][y] + a[x1][y1] == 3) return void(ans++);
if (a[x][y] == a[x1][y1] && a[x][y]) return;
if (a[x][y] == a[x1][y1] && !a[x][y]) return add(k1,k2,1),add(k2,k1,1),void();
add(k1,k2,1),add(k2,k1,1);
}
int main()
{
scanf("%d%d",&n,&m);
S = n * m + 1,T = S + 1;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
{
scanf("%d",&a[i][j]);
if (a[i][j] == 1) add(S,(i - 1) * m + j,inf);
if (a[i][j] == 2) add((i - 1) * m + j,T,inf);
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
{
int x1 = i - 1,y1 = j,x2 = i,y2 = j - 1;
if (x1 > 0) work(i,j,x1,y1);
if (y2 > 0) work(i,j,x2,y2);
}
while (bfs()) ans += dinic(S,inf,S);
printf("%d",ans);
}