POJ 3274 Gold Balanced Lineup

Gold Balanced Lineup

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 14378 Accepted: 4165

Description

Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities(类似). In fact, FJ has been able to narrow down the list of features(特色) shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For example, cows exhibiting(展览) feature #1 might have spots, cows exhibiting feature #2 might prefer C to Pascal, and so on.

FJ has even devised(设计) a concise(简明的) way to describe each cow in terms of its "feature ID", a single K-bit integer(整数) whose binary(二进制的) representation(代表) 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(敏感的) 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(连续的) 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

Line 1: Two space-separated integers(整数), N and K.
Lines 2..N+1: Line i+1 contains a single K-bit integer specifying(指定) the features(特色) present in cow i. The least-significant bit of this integer is 1 if the cow exhibits(展览品) 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(连续的) balanced group of cows.

Sample Input

7 37672142

Sample Output

4

Hint

In the range from cow #3 to cow #6 (of size 4), each feature(特色) appears in exactly 2 cows in this range

用vector一直在超时，无奈换成了链式前向星，然后就A了。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<cmath>using namespace std;int a[112345][35];struct node{    int num;    int next;}b[112345];int hs[112345];int ans;int m,k;int  in =99991;void push(int key,int i){    b[ans].num=i;    b[ans].next=hs[key];    hs[key]=ans;    ans++;}void ck(int key,int i){    int you=hs[key],flag=1;    while(you!=-1)    {        int tt=1;        for(int j=2;j<=k;j++)        {            if(a[b[you].num][j]!=a[i][j])            {                tt=0;                break;            }        }        if(tt)        {            m=max(m,(i-b[you].num));        }        you=b[you].next;    }    if(flag) push(key,i);}int main(){    int n;    cin>>n>>k;    ans=0;    memset(a,0,sizeof(a));    for(int i=0;i<=99992;i++) hs[i]=-1;    for(int i=1;i<=n;i++)    {        int x;        scanf("%d",&x);        int top=k;        while(x>0)        {            a[i][top--]=x%2;            x/=2;        }    }    for(int i=2;i<=n;i++)    for(int j=1;j<=k;j++) a[i][j]=a[i][j]+a[i-1][j];    for(int i=1;i<=n;i++)    for(int j=2;j<=k;j++) a[i][j]=a[i][j]-a[i][1];    for(int i=1;i<=k;i++) a[0][i]=0;    push(0,0);    m=0;    for(int i=1;i<=n;i++)    {        int key=0;        for(int j=2;j<=k;j++) key+=a[i][j];        key=abs(key)%in;        if(hs[key]==-1) push(key,i);        ck(key,i);    }    cout<<m<<endl;    return 0;}