Educational Codeforces Round 11C. Hard Process two pointer

地址：http://codeforces.com/contest/660/problem/C

题目：

You are given an array a with n elements. Each element of a is either 0 or 1.

Let's denote the length of the longest subsegment of consecutive elements in a, consisting of only numbers one, as f(a). You can change no more than k zeroes to ones to maximize f(a).

The first line contains two integers n and k (1 ≤ n ≤ 3·105, 0 ≤ k ≤ n) — the number of elements in a and the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 1) — the elements of a.

On the first line print a non-negative integer z — the maximal value of f(a) after no more than k changes of zeroes to ones.

On the second line print n integers aj — the elements of the array a after the changes.

If there are multiple answers, you can print any one of them.

7 1
1 0 0 1 1 0 1

4
1 0 0 1 1 1 1

10 2
1 0 0 1 0 1 0 1 0 1

5
1 0 0 1 1 1 1 1 0 1

 思路：一开始我用的是n^2的算法，一直tle，后来才知道有种算法叫尺取法：就是动态维护一个长度为x的区间，并同时记录起始位置和终点位置。

　　对这题而言，就是维护含0数为k的0,1区间，记录长度最大值，和起始位置和终点位置;

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <vector>

#define PI acos((double)-1)
#define E exp(double(1))
using namespace std;
int a[1000000];
int main (void)
{
int n,k,s=0,e=0,sum=0,len=0;
cin>>n>>k;
for(int i = 1; i<=n; i++)
{
scanf("%d",&a[i]);
sum += (a[i] == 0);
while(sum > k)
{
sum -= (a[++s] == 0);
}
if(len < i - s)
{
e = i;
len = i - s;
}
}
cout<<len<<endl;
for(int i = 1; i<=n; i++)
if(e>= i && i> e - len )
{
printf("1 ");
}
else
{
printf("%d ",a[i]);
}
return 0;
}