蓝桥杯备赛打卡Day2

1.复习了Acwing算法基础课第二章的三道题
(1)单链表
(2)滑动窗口
(3)KMP字符串
要点写在注释里面
2.独立完成蓝桥杯每日一题 5407.管道
明天再看题解

//滑动窗口
const int N = 1000010;
int a[N], n, k;
deque<int> dq;

int main()
{
	ios::sync_with_stdio(false);
	cin >> n >> k;
	for (int i = 0; i < n; i++) cin >> a[i];

	for (int i = 0; i < n; i++)
	{
	    //以下是最难理解的一步:比较队尾和清除比当前元素小(可以不取等)的队尾元素
	    //记忆:比尾去尾
	    //理解:队头至队尾有三个元素a1,a2,a3(显然a1<a2<a3),若要使当前元素a4能有机会成为队头的最小元素,必须清除障碍
		while (dq.size() && a[dq.back()] >= a[i]) dq.pop_back();
		dq.push_back(i);

		if (i >= k && dq.back() - dq.front() >= k) dq.pop_front();//注意不是dq.size()>=k
		if (i >= k - 1) cout << a[dq.front()] << " ";
	}

	cout << endl;
	dq.clear();

	for (int i = 0; i < n; i++)
	{
		while (dq.size() && a[dq.back()] <= a[i]) dq.pop_back();
		dq.push_back(i);

		if (i >= k && dq.back()-dq.front() >= k) dq.pop_front();
		if (i >= k - 1) cout << a[dq.front()] << " ";
	}

	return 0;
}

//KMP字符串
const int N = 1000010;
int n, m;
char s1[N], s2[N];
int ne[N];

int main()
{
	ios::sync_with_stdio(false);
	cin >> n >> s1 + 1;
	cin >> m >> s2 + 1;

	//样例:a b a b c
	//    0 0 1 2 0
	for (int i = 2, j = 0; i <= n; i++)
	{
		while (j && s1[j + 1] != s1[i]) j = ne[j];

		if (s1[j + 1] == s1[i]) j++;

		ne[i] = j;
	}

	//样例:a b a b a b c a
	for (int i = 1, j = 0; i <= m; i++)
	{
		while (j && s1[j + 1] != s2[i]) j = ne[j];

		if (s1[j + 1] == s2[i]) j++;

		if (j == n)
		{
			cout << i - n << ' ';
			j = ne[j];
		}
	}

	return 0;
}

//管道
typedef long long ll;
const ll N = 100010;
typedef pair<ll, ll> PII;
ll n, len;
PII a[N];//存储,a[i]={Li,Si}表示位于第Li段管道中央的阀门会在Si时刻打开

ll fun(ll t)
{
	//将a中的数据转换成区间,存入v,便于后面进行区间合并
	vector<PII> v;
	for (ll i = 0; i < n; i++)
	{
		ll l = a[i].first, s = a[i].second;
		if (s > t) continue;//注意!!若此段的放水时刻大于查询时刻,说明还没有放水
		ll left = l - (t - s), right = l + (t - s);
		v.push_back({ left,right });
	}

	sort(v.begin(), v.end());


	vector<PII>::iterator it = v.begin();
	ll pre_left = it->first, pre_right = it->second;
	if (pre_left > 1) return false;//返回false(1)

	for (vector<PII>::iterator it = v.begin() + 1; it != v.end(); it++)
	{
		ll left = it->first, right = it->second;

		//最理想的状态:pre_right会有改变
		if (left <= pre_right + 1 && right>=pre_right+1)
		{
			pre_right = right;
		}
		else if (left > pre_right + 1)
		{
			return false;//返回false(2)
		}
	}

	if (pre_right < len) return false;//返回false(3)

	return true;
}

int main()
{
	ios::sync_with_stdio(false);
	cin >> n >> len;

	//存储
	for (ll i = 0; i < n; i++) {
		ll l, s; cin >> l >> s;
		a[i] = { l,s };
	}

	//对时刻进行二分查找
	ll l = 1, r = 2e9 + 20;//注意边界

	while (l < r)
	{
		ll mid = l + r >> 1;
		if (fun(mid)) r = mid;
		else l = mid + 1;
	}

	cout << l << endl;
	
	return 0;
}