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[0..1000010] of longint;
var f:array[0..1,0..1000010] 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 2];
      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-1;
        end;
      until i>j;
      if l<j then sort(l,j);
      if i<r then sort(i,r);
    end;

  begin
    sort(1,m);
  end;

begin
  readln(m,n);
  for i:=1 to m do
    readln(b[i]);
  for i:=1 to n do
    readln(a[i]);
  qsort(a,n);
  qsort(b,m);
  p:=1;
  for i:=1 to n do
  begin
    while (p<=m) and (a[i]>=b[p]) do inc(p);
    c[i]:=p;
  end;
  k:=0;
  for i:=1 to n do
  begin
    k:=1-k;
    l:=max(i-1,c[i-1]-n);
    r:=min(m,c[i-1]+n);
    s:=max(i,c[i]-n);
    e:=min(m,c[i]+n);
    if i=1 then r:=m-n+1;
    f[k,s-1]:=2147483647;
    p:=l;
    for j:=s to e do
    begin
      f[k,j]:=f[k,j-1];
      while (p<j) and (p<=r) do
      begin
        f[k,j]:=min(f[k,j],f[1-k,p]);
        inc(p);
      end;
    end;
    for j:=s to e do
      if f[k,j]<>2147483647 then inc(f[k,j],abs(a[i]-b[j]));
  end;
  ans:=2147483647;
  for i:=s to e do
    ans:=min(ans,f[k,i]);
  writeln(ans);
end.