记横联通是一块横着的没有硬石头的地，把他们编号。竖联通同理。
对于一个空地，将其横联通编号和竖联通编号连边，二分图匹配，最大匹配为答案。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
int n, m, hr[55][55], sr[55][55], hcnt, scnt, hea[5005], cnt, yy, tt, lev[5005];
int maxFlow, cur[5005];
const int oo=0x3f3f3f3f;
char ss[55][55];
queue<int> d;
struct Edge{
int too, nxt, val;
}edge[10005];
void add_edge(int fro, int too, int val){
edge[cnt].nxt = hea[fro];
edge[cnt].val = val;
edge[cnt].too = too;
hea[fro] = cnt++;
}
void addEdge(int fro, int too, int val){
add_edge(fro, too, val);
add_edge(too, fro, 0);
}
bool bfs(){
memset(lev, 0, sizeof(lev));
lev[yy] = 1;
d.push(yy);
while(!d.empty()){
int x=d.front();
d.pop();
for(int i=hea[x]; i!=-1; i=edge[i].nxt){
int t=edge[i].too;
if(!lev[t] && edge[i].val>0){
lev[t] = lev[x] + 1;
d.push(t);
}
}
}
return lev[tt]!=0;
}
int dfs(int x, int lim){
if(x==tt) return lim;
int addFlow=0;
for(int &i=cur[x]; i!=-1; i=edge[i].nxt){
int t=edge[i].too;
if(lev[t]==lev[x]+1 && edge[i].val){
int tmp=dfs(t, min(lim-addFlow, edge[i].val));
edge[i].val -= tmp;
edge[i^1].val += tmp;
addFlow += tmp;
if(addFlow==lim) break;
}
}
return addFlow;
}
void dinic(){
while(bfs()){
for(int i=yy; i<=tt; i++) cur[i] = hea[i];
maxFlow += dfs(yy, oo);
}
}
int main(){
cin>>n>>m;
for(int i=1; i<=n; i++)
scanf("%s", ss[i]+1);
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++)
if(!hr[i][j] && ss[i][j]!='#'){
hcnt++;
for(int k=j; k<=m && ss[i][k]!='#'; k++)
hr[i][k] = hcnt;
}
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++)
if(!sr[i][j] && ss[i][j]!='#'){
scnt++;
for(int k=i; k<=n && ss[k][j]!='#'; k++)
sr[k][j] = scnt;
}
memset(hea, -1, sizeof(hea));
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++)
if(ss[i][j]=='*')
addEdge(hr[i][j], hcnt+sr[i][j], 1);
yy = 0; tt = hcnt + scnt + 1;
for(int i=1; i<=hcnt; i++)
addEdge(yy, i, 1);
for(int i=1; i<=scnt; i++)
addEdge(i+hcnt, tt, 1);
dinic();
cout<<maxFlow<<endl;
return 0;
}