【CF725D】Contest Balloons（贪心，堆）

题意：acm队伍可以得气球，相同气球数是一个排名。每个队伍有一个气球数上限，如果该队伍的气球数大于上限

        该队伍被淘汰。给了你队伍的气球数，你的气球可以给别人，问你最大可能的排名。

        (2 ≤ n ≤ 300 000) (0 ≤ ti ≤ wi ≤ 10^18)

思路：对每个初始t[i]>t[1]的i，将w[i]-t[i]+1放入小根堆中

        开始给气球，将给出气球后大于当前气球数的组的w[i]-t[i]+1继续放入堆中

        记录给后名次的最小值即可

const oo=2000000000000000000;
var q,a,b,c,d:array[1..310000]of int64;

    n,m,i,ans,t,w:longint;
    s:int64;

procedure swap(var x,y:int64);
var t:int64;
begin
 t:=x; x:=y; y:=t;
end;

procedure qsort(l,r:longint);
var i,j:longint;
    mid1,mid2,mid3:int64;
begin
 i:=l; j:=r; mid1:=a[(l+r)>>1]; mid2:=b[(l+r)>>1]; mid3:=c[(l+r)>>1];
 repeat
  while (mid1<a[i])or(mid1=a[i])and(mid2>b[i])or
        (mid1=a[i])and(mid2=b[i])and(mid3>c[i]) do inc(i);
  while (mid1>a[j])or(mid1=a[j])and(mid2<b[j])or
        (mid1=a[j])and(mid2=b[j])and(mid3<c[j]) do dec(j);
  if 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 qsort(l,j);
 if i<r then qsort(i,r);
end;

function min(x,y:longint):longint;
begin
 if x<y then exit(x);
 exit(y);
end;

procedure shiftup(x,m:longint);
begin
 while (x>1)and(q[x]<q[x>>1]) do
 begin
  swap(q[x],q[x>>1]);
  x:=x>>1;
 end;
end;

procedure shiftdown(x,m:longint);
var t:longint;
begin
 while (x<<1)<=m do
 begin
  t:=x<<1;
  if (t+1<=m)and(q[t]>q[t+1]) then t:=t+1;
  if q[x]>q[t] then
  begin
   swap(q[x],q[t]);
   x:=t;
  end
   else break;
 end;
end;

begin
 //assign(input,'cf725D.in'); reset(input);
 //assign(output,'cf725D.out'); rewrite(output);
 readln(n);
 for i:=1 to n do
 begin
  read(a[i],b[i]);
  c[i]:=i;
 end;
 qsort(1,n);
 for i:=1 to n do d[c[i]]:=i;
 for i:=1 to d[1]-1 do
 begin
  inc(m); q[m]:=b[i]-a[i]+1; shiftup(m,m);
 end;
 s:=a[d[1]];
 for i:=1 to d[1] do
  if a[i]>a[d[1]] then inc(ans);
 inc(ans); t:=ans;
 w:=d[1]+1;
 while m>0 do
 begin
  if s-q[1]>=0 then
  begin
   s:=s-q[1]; q[1]:=oo; shiftdown(1,m); dec(t);
   while (w<=n)and(a[w]>s) do
   begin
    inc(m); q[m]:=b[w]-a[w]+1; shiftup(m,m);
    inc(w); inc(t);
   end;
   ans:=min(ans,t);
  end
   else break;
 end;
 writeln(ans);
 //close(input);
 //close(output);
end.