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
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
Source
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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;
}