2019 Hang electric multi-school training 1012

 
Meaning of the questions: given n, c, k.
Then gives the n numbers (between 1 ~ c), Q longest interval length. The section numbers that satisfy any occurrence of k appears at least once.

Ideas:
we can probably meet the criteria to judge the interval, and then press the interval does not meet the index number of conditions for inter-partition again, to judge again.

Until all possible to meet the conditions of the interval are judged finished.

: Pit
under the subject of the relationship between the number of paying particular attention in the inter-partition, for treatment of more complicated boundary, (l, r) and the conditions are not satisfied.

#include<bits/stdc++.h>
using namespace std;
#define met(a,b) memset(a,b,sizeof(a))
int a[100009];
int ans=0;
map<int, vector<int> >pos;
int n,c,k;
void dfs(int l,int r)
{
    //cout<<l<<"............."<<r<<endl;
    if(r-l+1<k) return;
    if(l==r)
    {
        if(k<=1)ans=max(ans,1);
        return ;
    }
    else if(r<l) return ;
    int i=0;
    for( i=l; i<=r; i++){
            int tl=lower_bound(pos[a[i]].begin(),pos[a[i]].end(),l)-pos[a[i]].begin();
            int tr=upper_bound(pos[a[i]].begin(),pos[a[i]].end(),r)-pos[a[i]].begin();

            if(tr-tl<k) break;
    }
    if(i==r+1) ans=max(ans,r-l+1);
    else
    {
        int st=l,j;
        for(j=0;j<pos[a[i]].size();j++)if(st<=pos[a[i]][j])break;
          for(;j<pos[a[i]].size()&&pos[a[i]][j]<r;j++)
          {
              dfs(st,pos[a[i]][j]-1);
              st=pos[a[i]][j]+1;
          }
          if(j<pos[a[i]].size())
                   dfs(st,r);
          else
          {
              dfs(pos[a[i]][j-1]+1,r);
          }

    }
    return ;
}
int main()
{

    while(cin>>n>>c>>k)
    {
         for(int i=1;i<=c;i++)pos[i].clear();
        met(a,0);
        ans=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            pos[a[i]].push_back(i);
        }
        // cout<<"1"<<"        "<<n<<endl;
        dfs(1,n);

        cost<<ans<<endl;
    }
    return 0;
}