清兵线（dp）

分析：假设自己一共杀死k个小兵，目前杀死了x个，那么收益为t*（k-x），为了方便计算x，把dist从小到大排序，f[i,j,1]表示杀了i到j段小兵，最后在位置i的最优解，f[i,j,2]表示杀了i到j段小兵，最后在位置j的最优解，f[i,j,1]=max{ f[i+1][j][1]+m-(dist[i+1]-dist[i])(k-j+i),f[i+1][j][2]+m-(dist[j]-dist[i])(k-j+i)}
最后考虑答案。枚举k，再枚举i，找最大的max｛fi[i,i+k-1], fj[i,i+k-1]｝

代码

const
  maxn=500;
var
  f:array[0..maxn,0..maxn,1..2] of longint;
  a:array[0..maxn] of longint;
  i,j,k,l,n,m,ans:longint;

function max(x,y,z:longint):longint;
begin
  max:=x;
  if y>max then max:=y;
  if z>max then max:=z;
end;

procedure sort;
var
  i,j:longint;
begin
  for i:=1 to n-1 do
    for j:=i+1 to n do
      if a[i]>a[j] then
        begin
          a[0]:=a[i];a[i]:=a[j];a[j]:=a[0];
        end;
end;

begin
  assign(input,'kill.in');reset(input);
  assign(output,'kill.out');rewrite(output);
  readln(n,m);
  for i:=1 to n do
    readln(a[i]);
  sort;
  for k:=1 to n do
    begin
      for i:=1 to n do
        begin
          f[i,i,1]:=m-abs(a[i])*k;
          f[i,i,2]:=f[i,i,1];
          ans:=max(ans,f[i,i,1],0);
        end;
      for l:=2 to k do
        for i:=1 to n-l+1 do
          begin
            j:=i+l-1;
            f[i,j,1]:=max(f[i+1,j,1]+m-(k-j+i)*(a[i+1]-a[i]),f[i+1,j,2]+m-(k-j+i)*(a[j]-a[i]),-maxlongint);
            f[i,j,2]:=max(f[i,j-1,1]+m-(k-j+i)*(a[j]-a[i]),f[i,j-1,2]+m-(k-j+i)*(a[j]-a[j-1]),-maxlongint);
          end;
      for i:=1 to n-k+1 do
        ans:=max(ans,f[i,i+k-1,1],f[i,i+k-1,2]);
    end;
  writeln(ans);
  close(input);close(output);
end.