poj1980

首先想到费用流，但m<=100000还是算了吧
那就感觉要用dp了，首先将a，b排序
贪心一下可知，a,b的配对肯定不可能出现交叉
这样就可以dp了，复杂度O(nm)还是过不去
在贪心一下会发现，对于a[i],它只可能在j的左右n的范围内匹配(b[j]是b序列中第一个大于a[i]的）
这样就O(n2)了

 type list=array[..] of longint;
 var f:array[..,..] of longint;
     c,a,b:list;
     ans,p,k,i,j,l,r,s,e,n,m:longint;

 function min(a,b:longint):longint;
   begin
     if a>b then exit(b) else exit(a);
   end;

 function max(a,b:longint):longint;
   begin
     if a>b then exit(a) else exit(b);
   end;

 procedure qsort(var a :list;m:longint);
   procedure sort(l,r: longint);
     var i,j,x,y: longint;
     begin
       i:=l;
       j:=r;
       x:=a[(l+r) div ];
       repeat
         while a[i]<x do inc(i);
         while x<a[j] do dec(j);
         if not(i>j) then
         begin
           y:=a[i];
           a[i]:=a[j];
           a[j]:=y;
           inc(i);
           j:=j-;
         end;
       until i>j;
       if l<j then sort(l,j);
       if i<r then sort(i,r);
     end;

   begin
     sort(,m);
   end;

 begin
   readln(m,n);
   for i:= to m do
     readln(b[i]);
   for i:= to n do
     readln(a[i]);
   qsort(a,n);
   qsort(b,m);
   p:=;
   for i:= to n do
   begin
     while (p<=m) and (a[i]>=b[p]) do inc(p);
     c[i]:=p;
   end;
   k:=;
   for i:= to n do
   begin
     k:=-k;
     l:=max(i-,c[i-]-n);
     r:=min(m,c[i-]+n);
     s:=max(i,c[i]-n);
     e:=min(m,c[i]+n);
     if i= then r:=m-n+;
     f[k,s-]:=;
     p:=l;
     for j:=s to e do
     begin
       f[k,j]:=f[k,j-];
       while (p<j) and (p<=r) do
       begin
         f[k,j]:=min(f[k,j],f[-k,p]);
         inc(p);
       end;
     end;
     for j:=s to e do
       if f[k,j]<> then inc(f[k,j],abs(a[i]-b[j]));
   end;
   ans:=;
   for i:=s to e do
     ans:=min(ans,f[k,i]);
   writeln(ans);
 end.