[Rqnoj-434]LIGNJA

LIGNJA
DP题，最大空闲等于N-最小花费，这个存储方式倒是挺特别。
代码：

type
    link=^node;
    node=record
               x,t:longint;
               next:link;
               end;
var
   g:array[1..10001] of link;
   f:array[1..10001] of longint;
   n,k:longint;
procedure connect(a,t:longint);
var
   p:link;
begin
     new(p);
     p^.x:=a+t; p^.t:=t;
     p^.next:=g[a]; g[a]:=p;
end;
procedure main;
var
   i,a,t:longint;
   p:link;
begin
     readln(n,k);
     for i:=1 to k do begin
         readln(a,t);
         connect(a,t);
     end;
     fillchar(f,sizeof(f),0);
     for i:=n downto 1 do
         if g[i]=nil then f[i]:=f[i+1]
         else begin
              f[i]:=maxlongint;
              p:=g[i];
              while p<>nil do begin
                    if p^.t+f[p^.x]<f[i] then f[i]:=p^.t+f[p^.x];
                    p:=p^.next;
              end;
         end;
     write(n-f[1]);
end;
begin
     main;
end.