BZOJ 1702 [Usaco2007 Mar]Gold Balanced Lineup 平衡的队列 stl--map

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

传送门

一道不错的题目，和bzoj4236JOIOJI类似。

首先用前缀和sum[color][i]记录颜色color从1~i的出现总次数。

然后假如说区间[L,R]是合法的，那么满足：

sum[1][R]-sum[1][L-1]=

sum[2][R]-sum[2][L-1]=

……

=sum[k][R]-sum[k][L-1]

对式子作一些变形，可以得到

sum[2][R]-sum[1][R]=sum[2][L-1]-sum[1][L-1]

sum[3][R]-sum[2][R]=sum[3][L-1]-sum[2][L-1]

……

那么只要枚举前缀和中，相邻颜色的差，

然后如果说对于(L-1)和R，(K-1)个差都相等那么区间[L,R]是合法的。

一开始想着排序但是实在麻烦……

所以直接就哈希了。。

把数字连起来然后放到map里面。

#include
using namespace std;
int read(){
	int x=0,f=1;char ch=getchar();
	while (ch<'0' || ch>'9'){if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
const int 
	N=100005,
	COL=35;
int n,K;
int sum[COL][N];
map hash;
map thash;
void get(int x,int ii){
	for (int i=0;i