题目就是问如何用最小的板覆盖所有的草地。可以横着放，也可以竖着放，允许一个草地放多个点。

建图方法就是 每个横向的草地作为X，纵向连续的草地作为Y.     X连接Y的边表示，  这里有他们的公共点。。

很显然，覆盖所有草地，就是覆盖所有的边 ，二分图中，最小点覆盖 = 最大匹配

= =其实如果存在一条边未被选中的节点覆盖，则必然存在一条对应的增广路径

//tpl
//ipqhjjybj_tpl.h
//header.h
#include <cstdio>
#include <cstdlib>
#include <map>
#include <set>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>

#define mp(x,y) make_pair(x,y)
#define pii pair<int,int>
#define pLL pair<long long ,long long>
#define rep(i,j,k) for(int i = j; i < k;i++)

using namespace std;

const int INF = 0x3f3f3f3f;

const int N = 500;
int g[N][N];
int cx[N],cy[N];
int mark[N];
int nx,ny;

int dfs(int u)  
{  
    rep(v,0,ny) 
    {  
        if(g[u][v]&&!mark[v])//u和v不要搞反了  
        {  
            mark[v]=1;  
            if(cy[v]==-1||dfs(cy[v]))  
            {  
                cx[u]=v;  
                cy[v]=u;  
                return 1;  
            }  
        }  
    }  
    return 0;  
}  
int maxmatch()  
{  
    int res=0;  
    memset(cx,-1,sizeof(cx));  
    memset(cy,-1,sizeof(cy));  
    rep(i,0,nx)
    {  
        if(cx[i]==-1)  
        {  
            memset(mark,0,sizeof(mark));  
            res+=dfs(i);  
        }  
    }  
	// rep(i,0,nx){
			// printf("cx[%d] = %d\n",i,cx[i]);
			// printf("cy[%d] = %d\n",i,cy[i]);
		// }
    return res;  
}  

int a[N][N],b[N][N];
char s[N][N];
int main(){
	int n,m;
	while(scanf("%d %d",&n,&m)!=EOF){
		memset(g,0,sizeof(g));
		rep(i,0,n) scanf("%s",s[i]);
		int cnt = 0;
		rep(i,0,n)
			rep(j,0,m){
				if(s[i][j]=='*'){
					if(i==0 || s[i-1][j]=='.')
						a[i][j] = cnt++;
					else a[i][j] = a[i-1][j];
				}
			}
		nx = cnt;
		cnt = 0;
		rep(i,0,n)
			rep(j,0,m){
				if(s[i][j] == '*'){
					if(j==0 || s[i][j-1]=='.')
						b[i][j] = cnt++;
					else b[i][j] = b[i][j-1];
					 g[a[i][j]][b[i][j]] = 1;
				}				
			}
		ny = cnt;
		// rep(i,0,nx){
			// rep(j,0,ny)
				// printf("g[%d][%d]=%d ",i,j,g[i][j]);
			// printf("\n");
		// }
		
		printf("%d\n",maxmatch());
	}
	return 0;
}

附上我的 最大流写法。。

//tpl
//ipqhjjybj_tpl.h
//header.h
#include <cstdio>
#include <cstdlib>
#include <map>
#include <set>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>

#define mp(x,y) make_pair(x,y)
#define pii pair<int,int>
#define pLL pair<long long ,long long>
#define rep(i,j,k) for(int i = j; i < k;i++)

using namespace std;

const int INF = 0x3f3f3f3f;


const int N = 1111;
int tot;
int s,t;
int sum;
struct node{
	int u,v,w,next;
	node(){}
	node(int _u,int _v,int _w,int _next){
		u=_u,v=_v,w=_w,next=_next;
	}
}edge[N*N];
int head[N],cur[N],dis[N];
int pre[N],gap[N],aug[N];
const int oo=0x3f3f3f;
void addEdge(int u,int v,int w){
	edge[tot]=node(u,v,w,head[u]);
	head[u]=tot++;
	edge[tot]=node(v,u,0,head[v]);
	head[v]=tot++;
}

int SAP(int s,int e,int n){
	int max_flow=0,v,u=s;
	int id,mindis;
	aug[s]=oo;
	pre[s]=-1;
	memset(dis,0,sizeof(dis));
	memset(gap,0,sizeof(gap));
	gap[0]=n;

	for(int i=0;i <= n;i++)
		cur[i]=head[i];

	while(dis[s]<n){
		if(u==e){
			max_flow += aug[e];
			for(v=pre[e]; v!=-1; v=pre[v]){
				int ed=cur[v];
				edge[ed].w -= aug[e];
				edge[ed^1].w += aug[e];
				aug[v]-=aug[e];
				if(edge[ed].w==0) u=v;
			}
		}
		bool flag=false;
		for(id=cur[u]; id!=-1;id=edge[id].next){
			v=edge[id].v;
			if(edge[id].w > 0 && dis[u]==dis[v]+1){
				flag=true;
				pre[v]=u;
				cur[u]=id;
				aug[v]=min(aug[u],edge[id].w);
				u=v;
				break;
			}
		}
		if(flag==false){
			if(--gap[dis[u]] == 0) break; 
			int mindis=n;
			for(id=head[u]; id!=-1; id=edge[id].next){
				v=edge[id].v;
				if(edge[id].w>0 && dis[v] < mindis){
					mindis = dis[v];
					cur[u]=id;
				}
			}
			dis[u] = mindis + 1;
			gap[dis[u]]++;
			if(u!=s)u=pre[u];
		}
	}
	return max_flow;
}

int a[N][N],b[N][N];
char ss[N][N];
int main(){
	int n,m;
	while(scanf("%d %d",&n,&m)!=EOF){
		tot=sum=s=0;
		int tc=0;
		memset(head,-1,sizeof(head));

		rep(i,0,n) scanf("%s",ss[i]);
		int cnt = 0;
		t = ++cnt;
		rep(i,0,n)
			rep(j,0,m){
				if(ss[i][j]=='*'){
					if(i==0 || ss[i-1][j]=='.')
						a[i][j] = ++cnt , addEdge(s,a[i][j],1);
					else a[i][j] = a[i-1][j];
				}
			}
	
		rep(i,0,n)
			rep(j,0,m){
				if(ss[i][j] == '*'){
					
					if(j==0 || ss[i][j-1]=='.')
						b[i][j] = ++cnt ,addEdge(b[i][j],t,1);
					else b[i][j] = b[i][j-1];
					//g[a[i][j]][b[i][j]] = 1;
					addEdge(a[i][j],b[i][j],1);
					
				}				
			}
		
		printf("%d\n",SAP(s,t,cnt+1));
	}
	return 0;
}