POJ 1018

题目大意：多组询问，问每组，要买n种零件，每种零件都有各自的m种选择，每种零件有两种属性：带宽b和价格p，求一种方案，有max（min（all_b）/sum(p)）

解：discuss各种奇葩，无聊回去研究一下。我的做法是按b值排序，看b最大可以选多少（用多少b以上的能把n种零件买齐），如果fin点价格相同，则需要特殊处理，交换那些key点（使零件都有的点），再贪心，min[i] = 从i买到n的最小价格，然后枚举价格，从1 to fin即可，其实挺乱的，还是见程序吧。

//Communication system
const
        maxlen=10001;
        inf='1.txt';
type
        data=record
          b, p, col: longint;
        end;
var
        a: array[0..maxlen]of data;
        min: array[0..maxlen]of longint;
        tmp: array[0..100]of longint;
        fin, tot, test, wugu, m, n: longint;
        ans: extended;
procedure qsort(b, e: longint);
var
        i, j, x: longint;
        k: data;
begin
  i := b; j := e; x := a[(i+j)>>1].b;
  repeat
    while a[i].b<x do inc(i);
    while a[j].b>x do dec(j);
    if i<=j then begin
      k := a[i]; a[i] := a[j]; a[j] := k;
      inc(i); dec(j);
    end;
  until i>j;
  if j>b then qsort(b, j);
  if i<e then qsort(i, e);
end;

procedure give(b, e: longint);
var
        s, i, j: longint;
        rec: array[0..100]of longint;
begin
  filldword(rec, sizeof(rec)>>2, maxlongint);
  for i := e to tot do
    with a[i] do
      if rec[col]>p then rec[col] := p;
  s := 0;
  for i := 0 to 100 do if rec[i]<>maxlongint then s := s + rec[i];
  for i := e downto b do
    with a[i] do begin
      if rec[col]>p then begin
        s := s - rec[col] + p;
        rec[col] := p;
      end;
      min[i] := s;
    end;
end;

function deal(x: longint): longint;
var
        i, j: longint;
        k: data;
begin
  i := x;
  while a[i].p=a[i+1].p do inc(i);
  j := i;
  while j>=x do begin
    dec(tmp[a[j].col]);
    if tmp[a[j].col]=0 then if j<i then begin
      k := a[i]; a[i] := a[j]; a[j] := k;
      dec(i);
    end;
    dec(j);
  end;
  deal := i;
end;

procedure init;
var
        s, i, j: longint;
        k: data;
begin
  fillchar(min, sizeof(min), 0);
  fillchar(tmp, sizeof(tmp), 0);
  tot := 0; ans := 0;
  readln(n);
  for i := 1 to n do begin
    read(m);
    for j := 1 to m do begin
      inc(tot);
      with a[tot] do begin
        read(b, p);
        col := i;
      end;
    end;
    readln;
  end;
  qsort(1, tot);
  s := n;
  for i := tot downto 1 do begin          {!!!}
    if tmp[a[i].col]=0 then begin
      dec(s);
    end;
    inc(tmp[a[i].col]);
    j := i;
    if s=0 then break;
  end;
  fin := deal(j);
  give(1, fin);
end;

procedure main;
var
        i: longint;
begin
  for i := 1 to fin do begin
    if a[i].b/min[i]>ans then ans := a[i].b/min[i];
  end;
end;

procedure print;
begin
  writeln(ans:0:3);
end;
begin
  assign(input,inf); reset(input);
  readln(wugu);
  for test := 1 to wugu do begin
    init;
    main;
    print;
  end;
end.