1702: [Usaco2007 Mar]Gold Balanced Lineup 平衡的隊(duì)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)頭牛，一共K(1<=K<=30)種特色，
每頭牛有多種特色，用二進(jìn)制01表示它的特色I(xiàn)D。比如特色I(xiàn)D為13(1101)，
則它有第1、3、4種特色。[i,j]段被稱為balanced當(dāng)且僅當(dāng)K種特色在[i,j]內(nèi)
擁有次數(shù)相同。求最大的[i,j]段長度。

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

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鳴謝fjxmyzwd

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Source

Gold

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題解：一開始狠狠的逗比了一下——一開始我看到了這題，想當(dāng)然認(rèn)為問題可以轉(zhuǎn)化為求最長的和為\( {2}^{M} - 1 \)的倍數(shù)的子段，結(jié)果狠狠的WA了TT。。。這種想法有個最典型的反例，那就是連續(xù)\( {2}^{M} - 1 \)個1，但是很明顯不符合題意

于是發(fā)現(xiàn)如果某段內(nèi)各個位相等的話，那么對于各個位的前綴和之差必然完全相等，其實(shí)我們也不必直接去求前綴和之差，直接可以用平衡樹進(jìn)行形態(tài)存儲——形態(tài)存儲指的是將各個位上的累加數(shù)字關(guān)于第一個元素進(jìn)行個相對化——比如(2,4,6)可以轉(zhuǎn)化為(0,2,4),而(4,6,8)也可以轉(zhuǎn)為(0,2,4)這樣如果兩個前綴和數(shù)組可以構(gòu)成形態(tài)相等的話，那就意味著中間這一段符合題目中所述的各個位累加和相等，于是用一顆平衡樹存儲即可，時間復(fù)雜度\( O\left(N M \log N \right) \)

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.

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轉(zhuǎn)載于:https://www.cnblogs.com/HansBug/p/4427891.html

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