\(Soution\)
把木板看成一个点,把 \(*\) 看成一条边,因为一个 \(*\) 最多对应两块木板,现在问题转化成了二分图上的最小点覆盖问题,众所周知二分图上最小点覆盖等于最大匹配
\(Code\)
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 2505;
int n,m,cnta,cntb,tot = 1,h[N << 1],dep[N << 1],q[N << 1],cur[N << 1],S,T;
char s[55][55];
struct nd{
int l,r,id;
}a[N],b[N];
struct edge{
int to,nxt,z;
}e[N * N];
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 sum = 0,flow = 0;
for (int &i = cur[u]; i; i = e[i].nxt)
{
int v = e[i].to;
if (v == fa || dep[v] != dep[u] + 1 || 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) continue;
}
return flow;
}
int main()
{
scanf("%d%d",&n,&m);
for (int i = 1; i <= n; i++)
{
scanf("%s",s[i] + 1);
if (s[i][1] != '.') a[++cnta] = nd{1,1,i};
for (int j = 2; j <= m; j++)
{
if (s[i][j - 1] == '.' && s[i][j] != '.') a[++cnta] = nd{j,j,i};
if (s[i][j] != '.') a[cnta].r = j;
}
}
for (int j = 1; j <= m; j++)
{
if (s[1][j] != '.') b[++cntb] = nd{1,1,j};
for (int i = 2; i <= n; i++)
{
if (s[i - 1][j] == '.' && s[i][j] != '.') b[++cntb] = nd{i,i,j};
if (s[i][j] != '.') b[cntb].r = i;
}
}
S = cnta + cntb + 1,T = S + 1;
for (int i = 1; i <= cnta; i++) add(S,i,1);
for (int i = 1; i <= cntb; i++) add(i + cnta,T,1);
for (int i = 1; i <= cnta; i++)
for (int j = 1; j <= cntb; j++)
{
if (b[j].id > a[i].r || b[j].id < a[i].l) continue;
if (a[i].id > b[j].r || a[i].id < b[j].l) continue;
add(i,j + cnta,1);
}
int ans = 0;
while (bfs()) ans += dinic(S,2147483647,S);
printf("%d\n",ans);
}