题目大意：有n只牛，每只牛有k个属性，接下来n个数字，每个数字的二进制位上的1和0分别表示某种属性的有或者无，然后一个特殊数列就是，一个区间内所有牛的各种属性的总和相等（有e种1属性 e种2属性and so on），问你这排牛的最长的特殊数列长度是多少。
思路：看上去像dp，但思路走不通，然后看网上大佬的思路，仿佛推开新世界的大门。
数组sum[i][j]表示从的1到i头cow属性j的和。所以题目要求等价为求满足
sum[i][0]-sum[j][0]==sum[i][1]-sum[j][1]==.....==sum[i][k-1]-sum[j][k-1] (j<i)
最大的i-j
将上式变换为
sum[i][1]-sum[i][0]==sum[j][1]-sum[j][0]
sum[i][2]-sum[i][0]==sum[j][2]-sum[j][0]
.
sum[i][k-1]-sum[i][0]==sum[j][k-1]-sum[j][0]
令C[i][l]=sum[i][l]-sum[i][0] (0<l<k)。
所以只需求满足C[i]==C[j] 中最大的i-j。举样例来说明一下：
x 属性 牛
7 1 1 1 1
6 0 1 1 2
7 1 1 1 3
2 0 1 0 4
1 1 0 0 5
4 0 0 1 6
2 0 1 0 7
按行累加得sum[i]：
1 1 1
1 2 2
2 3 3
2 4 3
3 4 3
3 4 4
3 5 4
都减去第一列得c[i]：
0 0 0
0 1 1
0 1 1
0 2 1
0 1 0
0 1 1
0 2 1
所以说 最大区间是 6-2 = 4这道题最主要是还是让我理解了哈希的用处，虽然怎么用还再摸索中，但也慢慢的推开了哈希的大门了吧，还有这个题意的转化也非常的重要。
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
using namespace std;
typedef pair<int,int> pii;
typedef long long ll;
const double PI=acos(-1.0);
int fact[10]= {1,1,2,6,24,120,720,5040,40320,362880};
const int maxn = 100005;
const int MAX = 100005;
const int mod = 1000000;
int hash[MAX*10];//hash表储存下标
int next[MAX*10];//next作为邻接表
int sum[MAX][35];//第 1 头牛到第 i 头的对应属性的和
int c[MAX][35];//存放每头牛属性 j与第一个属性的差
int n,k;
int gethash(int *cc){
int key=0;
for(int i=1;i<=k;i++){
key=(key)%mod+cc[i]%mod*cc[i]%mod;//网上看的哈希函数
key%=mod;
}
return abs(key);//key可能是负数
}
bool compare(int id,int x){
for(int i=1;i<=k;i++){
if(c[id][i]!=c[x][i])return false;
}
return true;
}
int main(){
cin>>n>>k;
memset(hash,-1,sizeof(hash));
hash[0]=0;//这两行很重要 因为有可能整个序列都是平衡的
next[0]=-1;//而序列表达是i-j 所以有一部分的牛的哈希可能是0
int maxlen=0;
for(int i=1;i<=n;i++){
int x;
scanf("%d",&x);
for(int j=1;j<=k;j++){
sum[i][j]+=sum[i-1][j]+x%2;
c[i][j]=sum[i][j]-sum[i][1];
x/=2;
}
int key=gethash(c[i]);
for(int j=hash[key];j!=-1;j=next[j]){
if(compare(i,j)){
maxlen=max(i-j,maxlen);
}
}
next[i]=hash[key];//邻接表
hash[key]=i;
}
cout<<maxlen<<endl;
}| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 16727 | Accepted: 4736 |
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.
Input
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
Sample Input
7 3 7 6 7 2 1 4 2
Sample Output
4
Hint