bzoj2794

这题我得到一个经验，bool型的dp一定要想办法把bool去掉来表示更多的东西（1933也是这个道理）

暴力大家都会，这里有两个限制条件

一个限制条件我们可以排序不断加入，另一个呢

我们可以用f[i]表示c[]的和等于i时，最小的b[]最大是多少

然后转移判断就很容易解决了

var sum,a,b,c:array[0..1010] of longint;
    w,s,h,f,d:array[0..1000010] of longint;
    ans:array[0..1000010] of boolean;
    mx,i,n,j,k,q: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 swap(var a,b:longint);
  var c:longint;
  begin
    c:=a;
    a:=b;
    b:=c;
  end;

procedure sorta(l,r:longint);
  var i,j,x:longint;
  begin
    i:=l;
    j:=r;
    x:=a[(l+r) shr 1];
    repeat
      while a[i]<x do inc(i);
      while x<a[j] do dec(j);
      if not(i>j) then
      begin
        swap(a[i],a[j]);
        swap(b[i],b[j]);
        swap(c[i],c[j]);
        inc(i);
        dec(j);
      end;
    until i>j;
    if l<j then sorta(l,j);
    if i<r then sorta(i,r);
  end;

procedure sortq(l,r:longint);
  var i,j,x:longint;
  begin
    i:=l;
    j:=r;
    x:=w[(l+r) shr 1];
    repeat
      while w[i]<x do inc(i);
      while x<w[j] do dec(j);
      if not(i>j) then
      begin
        swap(w[i],w[j]);
        swap(s[i],s[j]);
        swap(h[i],h[j]);
        swap(d[i],d[j]);
        inc(i);
        dec(j);
      end;
    until i>j;
    if l<j then sortq(l,j);
    if i<r then sortq(i,r);
  end;

begin
  readln(n);
  for i:=1 to n do
    readln(c[i],a[i],b[i]);
  sorta(1,n);
  for i:=1 to n do
    sum[i]:=sum[i-1]+c[i];

  readln(q);
  for i:=1 to q do
  begin
    readln(w[i],h[i],s[i]);
    d[i]:=i;
    mx:=max(mx,h[i]);
  end;
  sortq(1,q);
  j:=1;
  f[0]:=1000000007;
  for i:=1 to q do
  begin
    while (j<=n) and (a[j]<=w[i]) do
    begin
      for k:=min(mx,sum[j]) downto c[j] do
        if f[k-c[j]]>0 then f[k]:=max(f[k],min(f[k-c[j]],b[j]));
      inc(j);
    end;
    ans[d[i]]:=(f[h[i]]>w[i]+s[i]);
  end;
  for i:=1 to q do
    if ans[i] then writeln('TAK') else writeln('NIE');
end.