可以从上下左右淹来，不能只b(0,0)
by bestzzy @ 2017-09-16 08:22:54
@[bestzzy](/space/show?uid=32854)
by 龙之吻—水货 @ 2017-09-19 20:27:08
我是先在周围添加一圈0 再进行搜索 所以相当于从四周淹过来吧？
by 龙之吻—水货 @ 2017-09-19 20:28:36
你应该只搜了左上，没搜右下，两边都得加，如果不是，我就不知道了；
~~我是P党~~，给下我的表
```cpp
var
f:array[1..4,1..2]of longint;
a:array[0..500,0..500]of longint;
b:array[0..500,0..500]of boolean;
i,j,n,m,ans:longint;
x:ansistring;
function check(x,y:longint):boolean;
begin
if (x>0)and(x<=n)and(y>0)and(y<=m)and(a[x,y]<>1)and(b[x,y]=true) then exit(true);
exit(false);
end;
procedure dfs(x,y:longint);
var i:longint;
begin
b[x,y]:=false;
for i:=1 to 4 do
if check(x+f[i,1],y+f[i,2]) then
dfs(x+f[i,1],y+f[i,2]);
exit;
end;
begin
readln(n,m);
f[1,1]:=0; f[1,2]:=1;
f[2,1]:=1; f[2,2]:=0;
f[3,1]:=0; f[3,2]:=-1;
f[4,1]:=-1; f[4,2]:=0;
for i:=1 to n do
begin
readln(x);
for j:=1 to length(x) do
if x[j]='*' then a[i,j]:=1;
end;
fillchar(b,sizeof(b),true);
for i:=1 to m do
begin
if a[1,i]=0 then
dfs(1,i);
if a[n,i]=0 then
dfs(n,i);
end;
for j:=1 to n do
begin
if a[j,1]=0 then
dfs(j,1);
if a[j,m]=0 then
dfs(j,m);
end;
for i:=1 to n do
for j:=1 to m do
if (b[i,j])and(a[i,j]=0) then inc(ans);
write(ans);
end.
```
by bestzzy @ 2017-09-21 17:15:34
数组开大点
by bestzzy @ 2017-09-21 17:16:32
橙名大佬竟会被这种题绊倒
by 宸之辰 @ 2017-09-21 23:29:12
@[bestzzy](/space/show?uid=32854) 这个数组是够用的吧（我开大了依然wa）
而且看评测信息的时候都是我输入的比答案小
by 龙之吻—水货 @ 2017-09-26 19:20:50
@[bestzzy](/space/show?uid=32854) 当我发现n，m都被我设置成字符，并改过来AC后，发现自己真的是蒟蒻了
by 龙之吻—水货 @ 2017-11-09 12:45:05
@[龙之吻—水货](/space/show?uid=49866) 666
by bestzzy @ 2017-11-09 16:29:37