思路
- 先按照颜色把每条狗存放起来
- 对它们的亲密度进行二分,求出第 k 对亲密关系的狗的亲密度是多少
- calc表示求小于等于x的有多少对数,我们利用异色点对数=总点对数-同色点对数来求
- 然后在枚举该其亲密度,把第 k 对亲密关系的狗输出
代码
// Problem: 温澈滢的狗狗
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/9984/D
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp(aa,bb) make_pair(aa,bb)
#define _for(i,b) for(int i=(0);i<(b);i++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,b,a) for(int i=(b);i>=(a);i--)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define X first
#define Y second
#define lowbit(a) (a&(-a))
#define debug(a) cout<<#a<<":"<<a<<"\n"
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef long double ld;
const int N=100010;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const double eps=1e-6;
const double PI=acos(-1.0);
int n,c[N];
ll k;
vector<int> g[N];
ll calc(ll x){
ll sum=x*(x-1)/2+x*(n-x);
for(int i=1;i<=n;i++){
int pos=0;
for(int j=0;j<g[i].size();j++){
while(pos<g[i].size()&&g[i][pos]-g[i][j]<=x) pos++;
sum-=pos-1-j;
}
}
return sum;
}
void solve(){
cin>>n>>k;
rep(i,1,n){
cin>>c[i];
g[c[i]].pb(i);
}
int l=1,r=n-1,ans=0;
while(l<=r){
int mid=(l+r)>>1;
if(calc(mid)>=k) r=mid-1,ans=mid;
else l=mid+1;
}
if(!ans){
cout<<"-1\n";
return;
}
ll kk=calc(ans-1);
rep(i,1,n-ans){
if(c[i]!=c[i+ans]) kk++;
if(kk==k){
cout<<i<<" "<<i+ans<<"\n";
return;
}
}
}
int main(){
ios::sync_with_stdio(0);cin.tie(0);
// int t;cin>>t;while(t--)
solve();
return 0;
}