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
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\( O\left(N M \log N \right) \)
1 /************************************************************** 2 Problem: 1702 3 User: HansBug 4 Language: Pascal 5 Result: Accepted 6 Time:1016 ms 7 Memory:13116 kb 8 ****************************************************************/ 9 10 type 11 list=array[1..30] of longint; 12 var 13 i,j,k,l,m,n,head:longint; 14 a:array[0..100005] of list; 15 fix,lef,rig:array[0..100005] of longint; 16 function putin(x:longint;var a:list):longint; 17 var i:longint; 18 begin 19 fillchar(a,sizeof(a),0); 20 i:=0; 21 while x>0 do 22 begin 23 inc(i); 24 a[i]:=x mod 2; 25 x:=x div 2; 26 end; 27 end; 28 function min(x,y:longint):longint; 29 begin 30 if x<y then min:=x else min:=y; 31 end; 32 function max(x,y:longint):longint; 33 begin 34 if x>y then max:=x else max:=y; 35 end; 36 function fc(a,b:list):longint; 37 var i,j,k:longint; 38 begin 39 fc:=0; 40 for i:=1 to m do 41 begin 42 j:=(a[i]-a[1])-(b[i]-b[1]); 43 if j>0 then exit(1); 44 if j<0 then exit(-1); 45 end; 46 end; 47 procedure lt(var x:longint); 48 var f,r:longint; 49 begin 50 if (x=0) or (rig[x]=0) then exit; 51 f:=x;r:=rig[x]; 52 rig[f]:=lef[r]; 53 lef[r]:=f; 54 x:=r; 55 end; 56 procedure rt(var x:longint); 57 var f,l:longint; 58 begin 59 if (x=0) or (lef[x]=0) then exit; 60 f:=x;l:=lef[x]; 61 lef[f]:=rig[l]; 62 rig[l]:=f; 63 x:=l; 64 end; 65 function ins(var x:longint;y:longint):longint; 66 begin 67 if x=0 then 68 begin 69 x:=y; 70 exit(y); 71 end; 72 j:=fc(a[x],a[y]); 73 case j of 74 0:exit(x); 75 1:begin 76 if lef[x]=0 then 77 begin 78 lef[x]:=y; 79 ins:=y; 80 end 81 else ins:=ins(lef[x],y); 82 if fix[lef[x]]<fix[x] then rt(x); 83 end; 84 -1:begin 85 if rig[x]=0 then 86 begin 87 rig[x]:=y; 88 ins:=y; 89 end 90 else ins:=ins(rig[x],y); 91 if fix[rig[x]]<fix[x] then lt(x); 92 end; 93 end; 94 end; 95 begin 96 readln(n,m);randomize; 97 fillchar(lef,sizeof(lef),0); 98 fillchar(rig,sizeof(rig),0); 99 for i:=1 to n+1 do 100 begin 101 if i=1 then putin(0,a[i]) else 102 begin 103 readln(j); 104 putin(j,a[i]); 105 end; 106 for j:=1 to m do a[i][j]:=a[i-1][j]+a[i][j]; 107 fix[i]:=random(maxlongint); 108 end; 109 head:=0;l:=0; 110 for i:=1 to n+1 do 111 begin 112 j:=ins(head,i); 113 l:=max(l,i-j); 114 end; 115 writeln(l); 116 readln; 117 end.