将植株按花生数从大到小排序,然后按排序后的顺序摘,每次摘前计算能否在摘后回到路边,如果能就将ans加上该植株花生数,如果不能就直接输出当前ans并退出。
var a:array[1..405,1..3] of integer;
i,j,x,y,m,n,k,n1,t1,ans,time:longint;
procedure sort(l,r:longint);
var i,j,m,t:longint;
begin
i:=l;j:=r;
m:=a[(l+r) div 2,1];
repeat
while a[i,1]>m do inc(i);
while a[j,1]<m do dec(j);
if (i<=j) then
begin
t:=a[i,1];a[i,1]:=a[j,1];a[j,1]:=t;
t:=a[i,2];a[i,2]:=a[j,2];a[j,2]:=t;
t:=a[i,3];a[i,3]:=a[j,3];a[j,3]:=t;
inc(i);dec(j);
end;
until i>j;
if l<j then sort(l,j);
if i<r then sort(i,r);
end;
begin
readln(m,n,k);
for i:=1 to m do
for j:=1 to n do
begin
read(x);
if x<>0
then
begin
inc(n1);
a[n1,1]:=x;
a[n1,2]:=i;
a[n1,3]:=j;
end;
end;
sort(1,n1);
x:=0;y:=a[1,3];
for i:=1 to n1 do
begin
t1:=time+abs(a[i,2]-x)+abs(a[i,3]-y)+1;
if t1+a[i,2]<=k
then
begin
time:=t1;
x:=a[i,2];
y:=a[i,3];
ans:=ans+a[i,1];
end
else
begin
writeln(ans);
exit;
end;
end;
writeln(ans);
end.