Educational Codeforces Round 11 C. Hard Process(two pointers)

题意:

$N\le 3\times 10^5的序列,A_i = 0 或者1$
$k\le N次操作将0变成1,求最长的连续1序列长度,并打印方案$

分析:

$two pointers经典题啦,记录下比较好的写法$
$时间复杂度O(n)$

代码:

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//
// Created by TaoSama on 2016-04-08
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 3e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, k, a[N];
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d%d", &n, &k) == 2) {
for(int i = 1; i <= n; ++i) scanf("%d", a + i);
int cnt = 0;
pair<int, int> ans = {0, 0}; //sum, l
for(int l = 1, r = 1; l <= n; ++l) {
while(r <= n && cnt + !a[r] <= k) cnt += !a[r++];
ans = max(ans, {r - l, l});
cnt -= !a[l];
}
printf("%d\n", ans.first);
for(int i = ans.second; ans.first--; ++i) a[i] = 1;
for(int i = 1; i <= n; ++i) printf("%d%c", a[i], " \n"[i == n]);
}
return 0;
}


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