1702: [Usaco2007 Mar]Gold Balanced Lineup 平衡的队列

1702: [Usaco2007 Mar]Gold Balanced Lineup 平衡的队列

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 510  Solved: 196
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Description

Farmer John's N cows (1 <= N <= 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 <= K <= 30). For example, cows exhibiting feature #1 might have spots, cows exhibiting feature #2 might prefer C to Pascal, and so on. FJ has even devised a concise way to describe each cow in terms of its "feature ID", a single K-bit integer whose binary representation tells us the set of features exhibited by the cow. As an example, suppose a cow has feature ID = 13. Since 13 written in binary is 1101, this means our cow exhibits features 1, 3, and 4 (reading right to left), but not feature 2. More generally, we find a 1 in the 2^(i-1) place if a cow exhibits feature i. Always the sensitive fellow, FJ lined up cows 1..N in a long row and noticed that certain ranges of cows are somewhat "balanced" in terms of the features the exhibit. A contiguous range of cows i..j is balanced if each of the K possible features is exhibited by the same number of cows in the range. FJ is curious as to the size of the largest balanced range of cows. See if you can determine it.

N(1<=N<=100000)头牛，一共K(1<=K<=30)种特色，
每头牛有多种特色，用二进制01表示它的特色ID。比如特色ID为13(1101)，
则它有第1、3、4种特色。[i,j]段被称为balanced当且仅当K种特色在[i,j]内
拥有次数相同。求最大的[i,j]段长度。

Input

* Line 1: Two space-separated integers, N and K.

* Lines 2..N+1: Line i+1 contains a single K-bit integer specifying the features present in cow i. The least-significant bit of this integer is 1 if the cow exhibits feature #1, and the most-significant bit is 1 if the cow exhibits feature #K.

Output

* Line 1: A single integer giving the size of the largest contiguous balanced group of cows.

Sample Input

7 3
7
6
7
2
1
4
2

INPUT DETAILS:

The line has 7 cows with 3 features; the table below summarizes the
correspondence:
Feature 3: 1 1 1 0 0 1 0
Feature 2: 1 1 1 1 0 0 1
Feature 1: 1 0 1 0 1 0 0
Key: 7 6 7 2 1 4 2
Cow #: 1 2 3 4 5 6 7

Sample Output

4

OUTPUT DETAILS:

In the range from cow #3 to cow #6 (of size 4), each feature appears
in exactly 2 cows in this range:
Feature 3: 1 0 0 1 -> two total
Feature 2: 1 1 0 0 -> two total
Feature 1: 1 0 1 0 -> two total
Key: 7 2 1 4
Cow #: 3 4 5 6

HINT

鸣谢fjxmyzwd

Source

Gold

题解：一开始狠狠的逗比了一下——一开始我看到了这题，想当然认为问题可以转化为求最长的和为\( {2}^{M} - 1 \)的倍数的子段，结果狠狠的WA了TT。。。这种想法有个最典型的反例，那就是连续\( {2}^{M} - 1 \)个1，但是很明显不符合题意

于是发现如果某段内各个位相等的话，那么对于各个位的前缀和之差必然完全相等，其实我们也不必直接去求前缀和之差，直接可以用平衡树进行形态存储——形态存储指的是将各个位上的累加数字关于第一个元素进行个相对化——比如(2,4,6)可以转化为(0,2,4),而(4,6,8)也可以转为(0,2,4)这样如果两个前缀和数组可以构成形态相等的话，那就意味着中间这一段符合题目中所述的各个位累加和相等，于是用一颗平衡树存储即可，时间复杂度\( O\left(N M \log N \right) \)

 /**************************************************************
     Problem:
     User: HansBug
     Language: Pascal
     Result: Accepted
     Time: ms
     Memory: kb
 ****************************************************************/

 type
     list=array[..] of longint;
 var
    i,j,k,l,m,n,head:longint;
    a:array[..] of list;
    fix,lef,rig:array[..] of longint;
 function putin(x:longint;var a:list):longint;
          var i:longint;
          begin
               fillchar(a,sizeof(a),);
               i:=;
               while x> do
                     begin
                          inc(i);
                          a[i]:=x mod ;
                          x:=x div ;
                     end;
          end;
 function min(x,y:longint):longint;
          begin
               if x<y then min:=x else min:=y;
          end;
 function max(x,y:longint):longint;
          begin
               if x>y then max:=x else max:=y;
          end;
 function fc(a,b:list):longint;
          var i,j,k:longint;
          begin
               fc:=;
               for i:= to m do
                   begin
                        j:=(a[i]-a[])-(b[i]-b[]);
                        if j> then exit();
                        if j< then exit(-);
                   end;
          end;
 procedure lt(var x:longint);
           var f,r:longint;
           begin
                if (x=) or (rig[x]=) then exit;
                f:=x;r:=rig[x];
                rig[f]:=lef[r];
                lef[r]:=f;
                x:=r;
           end;
 procedure rt(var x:longint);
           var f,l:longint;
           begin
                if (x=) or (lef[x]=) then exit;
                f:=x;l:=lef[x];
                lef[f]:=rig[l];
                rig[l]:=f;
                x:=l;
           end;
 function ins(var x:longint;y:longint):longint;
          begin
               if x= then
                  begin
                       x:=y;
                       exit(y);
                  end;
               j:=fc(a[x],a[y]);
               case j of
                    :exit(x);
                    :begin
                           if lef[x]= then
                              begin
                                   lef[x]:=y;
                                   ins:=y;
                              end
                           else ins:=ins(lef[x],y);
                           if fix[lef[x]]<fix[x] then rt(x);
                    end;
                    -:begin
                            if rig[x]= then
                               begin
                                    rig[x]:=y;
                                    ins:=y;
                               end
                            else ins:=ins(rig[x],y);
                            if fix[rig[x]]<fix[x] then lt(x);
                    end;
               end;
          end;
 begin
      readln(n,m);randomize;
      fillchar(lef,sizeof(lef),);
      fillchar(rig,sizeof(rig),);
      for i:= to n+ do
          begin
               if i= then putin(,a[i]) else
                  begin
                       readln(j);
                       putin(j,a[i]);
                  end;
               for j:= to m do a[i][j]:=a[i-][j]+a[i][j];
               fix[i]:=random(maxlongint);
          end;
      head:=;l:=;
      for i:= to n+ do
          begin
               j:=ins(head,i);
               l:=max(l,i-j);
          end;
      writeln(l);
      readln;
 end.