codeforces 616D Longest k-Good Segment(two pointer)
2016-01-13 14:03
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D. Longest k-Good Segment
The array a with n integers is given. Let's call the sequence of one or more consecutive elements in a segment. Also let's call the segment k-good if it contains no more than k different values.Find any longest k-good segment.
As the input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printfinstead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input
The first line contains two integers n, k (1 ≤ k ≤ n ≤ 5·105) — the number of elements in a and the parameter k.
The second line contains n integers ai (0 ≤ ai ≤ 106) — the elements of the array a.
Output
Print two integers l, r (1 ≤ l ≤ r ≤ n) — the index of the left and the index of the right ends of some k-good longest segment. If there are several longest segments you can print any of them. The elements in a are numbered from 1 to n from left to right.
Sample test(s)
input
5 5 1 2 3 4 5
output
1 5
input
9 3 6 5 1 2 3 2 1 4 5
output
3 7
input
3 1 1 2 3
output
1 1
//从左到右维护一个最长的不同元素个数<=k的区间 #include<cstdio> #include<cstring> #include<stack> #include<iterator> #include<queue> #include<set> #include<vector> #include<iostream> #include<map> #include<string> #include<algorithm> using namespace std; typedef long long ll; typedef unsigned long long ull; const int INF=0x3f3f3f3f; const int maxn=5e5+5; int cnt[1000005]; int a[maxn]; int main() { int n,k; scanf("%d%d",&n,&k); for(int i=0;i<n;i++) scanf("%d",&a[i]); int p1=0,p2=-1,l=0,r=0,tot=0; while(p2<n) { while(++p2<n) { if(!cnt[a[p2]])tot++; cnt[a[p2]]++; if(tot>k)break; if(p2-p1>=r-l)r=p2,l=p1; } while(tot>k) { cnt[a[p1]]--; if(cnt[a[p1]]==0) { tot--;p1++; break; } p1++; } } printf("%d %d\n",l+1,r+1); return 0; }
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- D. Longest k-Good Segment