ZOJ 3790 Consecutive Blocks（尺取法）

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

题意：求在n个数的数列中最多去掉k个数能得到的连续相同数字的最大数量。

思路：这题有很多方法，刚学了尺取法，我就用尺取法做，其中循环结束条件应为是s == n - 1 && t == n，开始我设为t == n,wa了几发。

#include<cstdio>#include<algorithm>#include<map>using namespace std;int a[100010];int main(){    int n, K;    while(~scanf("%d%d", &n, &K))    {        map<int, int> coun;        for(int i = 0; i < n; i++)        {            scanf("%d", &a[i]);        }        int res = 0, s = 0, t = 0, k = 0;        for(;;)        {            while(k < K && t < n )            {                if(a[t] != a[s])                {                    k++;                    coun[a[t++]]++;                }                else                {                    coun[a[t++]]++;                }            }            while(k == K && t < n && a[t] == a[s])            {                coun[a[t++]]++;            }            res = max(res, coun[a[s]]);            if(s == n - 1 && t == n) break; //结束条件应为 s == n - 1 && t == n 而不是只有 t == n;            if(a[s] == a[s + 1])            {                coun[a[s]]--;                s++;            }            else            {                coun[a[s]]--;                s++;                k = (t - s) - coun[a[s]];            }        }        printf("%d\n", res);    }    return 0;}