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.
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.
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.
5 5 1 2 3 4 5
1 5
9 3 6 5 1 2 3 2 1 4 5
3 7
3 1 1 2 3
1 1
two pointers一种不错的编程技巧。
从前往后,或者从两头向中间,可使算法时间复杂度大大降低。
题目要求从所给序列找到一段尽量长的连续系列,使得这段序列不同数字个数(cnt)不大于K。
一个暴力的思路是:对于原始序列枚举每个点作为左端点,找到对应最大的右端点。记为(Li,Ri)
假设从左向右边枚举,当枚举Li+1时,先处理Li位置,然后,明显当前Ri<=Ri+1;从Ri+1开始修改cnt值就可以了。
最后取最大值(Ri-Li)。
#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<set> #include<vector> #include<time.h> #include<string.h> using namespace std; #define LL __int64 #define N 1000005 int a[N],num[N]; int main() { int i,n,k; int l,r,cnt,pos; while(~scanf("%d%d",&n,&k)){ for(i=1;i<=n;++i){ scanf("%d",&a[i]); } l=r=1; cnt=1; pos=2; memset(num,0,sizeof(num)); num[a[1]]=1; for(i=1;i<=n;++i){ while(pos<=n&&cnt<=k){ ++num[a[pos]]; if(num[a[pos]]==1){ ++cnt; } if(cnt<=k) ++pos; } if(r-l<pos-1-i){ r=pos-1; l=i; } --num[a[i]]; if(num[a[i]]==0){ --cnt; ++pos; } } printf("%d %d\n",l,r); } return 0; }