POJ 3274 Gold Balanced Lineup

Gold Balanced Lineup

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 9652		Accepted: 2881

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

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

Sample Output

4

Hint

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

Source

USACO 2007 March Gold

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做完这道题我叹服他的思路，推荐大家看的博客：http://blog.youkuaiyun.com/hqd_acm/article/details/5902792

我有点晕，本以为我写的这题会超时，结果却没有超时，我是在输入的过程中，就所有的数都放在哈希表中，最后统一进行处理，本来以为会超时的结果却发现没有超时。

#include <stdio.h>
#include <string.h>
int sum[100010][31];
int c[100010][31];
struct link
{
    int val[31],next,pos;
}a[1000000];
int b[1000004],statck[1000000];
int maxlen=1000003;
int maxres,INF=0x7fffffff;
int main()
{
    int i,j,n,m,x,s,top,s1,s2,top2,u,v,max;
    scanf("%d %d",&n,&m);
    memset(sum,0,sizeof(sum));
    memset(b,-1,sizeof(b));
    memset(c,0,sizeof(c));
    top=0; top2=0;
    a[top].pos=0;
    for(i=1;i<=m;i++)
    {
        a[top].val[i]=0;
    }
    a[top].next=b[0]; b[0]=top; top++;
    statck[top2++]=0;
    for(i=1;i<=n;i++)
    {
        scanf("%d",&x);
        for(j=1;j<=m;j++)
        {
            sum[i][j]=sum[i-1][j]+x%2;
            c[i][j]=sum[i][j]-sum[i][1];
            x=x/2;
        }
        for(j=1,s=0;j<=m;j++)
        {
            s=(s*10%maxlen+c[i][j])%maxlen;
        }
        s=abs(s);
        if(b[s]==-1)
        {
            statck[top2++]=s;
        }
        for(j=1;j<=m;j++)
        {
            a[top].val[j]=c[i][j];
        }
        a[top].pos=i;
        a[top].next=b[s]; b[s]=top;
        top++;
    }
    max=0;
    for(i=0;i<=top2-1;i++)
    {
        s=statck[i];
        for(j=b[s];j!=-1;j=a[j].next)
        {
            if(a[j].pos!=INF)
            {
                for(u=a[j].next;u!=-1;u=a[u].next)
                {
                    for(v=1;v<=m;v++)
                    {
                        if(a[j].val[v]!=a[u].val[v])
                        {
                            break;
                        }
                    }
                    if(v==m+1)
                    {
                        if(abs(a[u].pos-a[j].pos)>max)
                        {
                            max=abs(a[u].pos-a[j].pos);
                        }
                        a[u].pos=INF;
                    }
                }
            }
        }
    }
    printf("%d\n",max);
    return 0;
}