ZOJ Consecutive Blocks（左右指针移动）

题意很简单就是让你看看隔着k个不同的，最多还能构成多少个连续的啊。

10^5托托的不能o(n^2)，显然是左右指针瞎搞啊，但是这道题可以排序使得更简单。

按照数字的大小，和地址的先后二级排序，然后分相同数字和不相同数字进行讨论，找到满足条件的最优的解。

Consecutive Blocks

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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There are N (1 ≤ N ≤ 10⁵) colored blocks (numbered 1 to N from left to right) which are lined up in a row. And the i-th block's color is C_i (1 ≤ C_i ≤ 10⁹). Now you can remove at most K (0 ≤ K ≤N) blocks, then rearrange the blocks by their index from left to right. Please figure out the length of the largest consecutive blocks with the same color in the new blocks created by doing this.

For example, one sequence is {1 1 1 2 2 3 2 2} and K=1. We can remove the 6-th block, then we will get sequence {1 1 1 2 2 2 2}. The length of the largest consecutive blocks with the same color is 4.

Input

Input will consist of multiple test cases and each case will consist of two lines. For each test case the program has to read the integers N and K, separated by a blank, from the first line. The color of the blocks will be given in the second line of the test case, separated by a blank. The i-th integer means C_i.

Output

Please output the corresponding length of the largest consecutive blocks, one line for one case.

Sample Input

8 11 1 1 2 2 3 2 2

Sample Output

4

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>#define eps 1e-8#define M 1000100//#define LL __int64#define LL long long#define INF 0x7ffffff#define PI 3.1415926535898const int maxn = 101000;using namespace std;struct node{    int x;    int point;}f[maxn];bool cmp(node a, node b){    if(a.x == b.x)        return a.point < b.point;    return a.x < b.x;}int main(){    int n, k;    while(cin >>n>>k)    {        int x;        for(int i = 0; i < n; i++)        {            cin >>x;            f[i].x = x;            f[i].point = i;        }        sort(f, f+n, cmp);        int Max = -1;        int i, j;        int r = 1;        int dis = 1;        for(i = 0; i < n;)        {            dis = r-i;            for(j = r; j < n; j++)            {                if(f[i].x != f[j].x)                {                    Max = max(Max, dis);                    i = j;                    r = j+1;                    break;                }                if(f[j].point-f[i].point > j-i+k)                {                    Max = max(Max, dis);                    i++;                    dis--;                    r = j;                    break;                }                dis++;            }            if(j == n)            {                Max = max(dis, Max);                break;            }        }        cout<<Max<<endl;    }    return 0;}