#include <iostream>
#include<cstdlib>
#include<cstdio>
#include<map>
#include<vector>
#include<cstring>
#include<algorithm>
#define FOR(i,x,y) for(i=x;i<=y;++i)
#define maxa 1000+100
using namespace std;
int n,m,ans = 0,b[maxa][maxa],cnt = 0,flag = 0;
string s[maxa];
int l1,r1,u,d;
int l[] = {0,0,1,-1};
int r[] = {1,-1,0,0};
void dfs(int x,int y)
{
    int i;
   FOR(i,0,3)
   {
       int nx = x+r[i];
       int ny = y+l[i];
       if(nx<0||nx>=n||ny<0||ny>=m)
        continue;
       if(s[nx][ny]=='#'&&b[nx][ny]==0){
        b[nx][ny] = 1;
        l1 = min(l1,ny);
        r1 = max(r1,ny);
        u = min(u,nx);
        d = max(d,nx);
         dfs(nx,ny);
       }
   }
}
bool check()
{
    int i,j;
    FOR(i,u,d)
    FOR(j,l1,r1)
    if(b[i][j]==0)
    return true;
    return false;
}
int main()
{
    int i,j;
    memset(b,0,sizeof(b));
    cin>>n>>m;
    FOR(i,0,n-1)
    {
        cin>>s[i];
    }
    FOR(i,0,n-1)
    FOR(j,0,m-1)
        if(s[i][j]=='#'&&b[i][j]==0)
        {
            cnt++;
            b[i][j] =1;
            l1 =r1 = j;
            u = d = i;
            dfs(i,j);
            if(check())
            {
                cout<<"Bad placement.";
                return 0;
            }
        }
    cout<<"There are "<<cnt<<" ships."<<endl;
    return 0;

}

也是以前做过的题呢
做法是找到‘#’进入搜索，把与它相连的所有‘#’都变成除了‘.’以外的字符
最重要的是搜索同时记录该船所能覆盖到的最远端点
即船的第一排与最后一排，第一列与最后一列，我这里用maxx minx maxy miny表示
从搜索中退出后循环这条船覆盖的区域，看有无‘.’（这里也就是为什么上面不变成‘.’）
有‘.’就bad placement 没有就加ans

var
    m:array[0..1001,0..1001] of char;
    i,j,k,L,ans,r,c,maxx,maxy,minx,miny:longint;
    

procedure se(i,j:longint);
begin
    if (i>maxx) then maxx:=i;
    if (i<minx) then minx:=i;
    if (j>maxy) then maxy:=j;
    if (j<miny) then miny:=j;
    m[i,j]:='@';
    if (m[i-1,j]='#') then se(i-1,j);
    if (m[i+1,j]='#') then se(i+1,j);
    if (m[i,j-1]='#') then se(i,j-1);
    if (m[i,j+1]='#') then se(i,j+1);//我比较喜欢拍机械键盘所以就这样多拍拍不用循环了
                                                                    //：P
end;
begin
    readln(r,c);
    for i:= 1 to r do
    begin
        for j:= 1 to c do
        begin
            read(m[i,j]);
        end;
        readln;
    end;
    for i:= 1 to r do
    begin
        for j:= 1 to c do
        begin
            if m[i,j]='#' then
            begin
                maxx:=-1;minx:=1001;
                maxy:=-1;miny:=1001;
                se(i,j);
                for k:= minx to maxx do
                begin
                    for L:= miny to maxy do
                    begin
                        if m[k,L]='.' then begin writeln('Bad placement.');halt; end;
                    end;
                end;
                inc(ans);
            end;
        end;
    end;
    writeln('There are ',ans,' ships.');
end.