P2251 质量检测(分块线段树RMQ单调队列)

P2251 质量检测
正解应该是ST表和单调队列，不过对于这道题来说只有查询没有修改，这里我还是想用线段树和分块来写，不得不说分块是真好，优雅的暴力
线段树版本：

#include  
#define LL long long 
using namespace std;

const int N = 1e6 + 10;

struct SegmentTree{
	int l, r;
	int minn;
	#define l(x) tree[x].l
	#define r(x) tree[x].r
	#define minn(x) tree[x].minn
}tree[4 * N]; 
int n, m, a[N];

void build(int p, int l, int r)
{
	l(p) = l, r(p) = r;
	if(l == r)	{minn(p) = a[l]; return;}
	int mid = (l + r) >> 1;
	build(p << 1, l, mid);
	build(p << 1 | 1, mid + 1, r);
	minn(p) = min(minn(p << 1), minn(p << 1 | 1));
}


int ask(int p, int l, int r)
{
	if(l <= l(p) && r >= r(p))	return minn(p);
	int mid = (l(p) + r(p)) >> 1;
	int val = 1e9;
	if(l <= mid) val = min(ask(p << 1, l , r), val);
	if(r > mid) val = min(ask(p << 1 | 1, l, r), val);
	return val;
}

void solve()
{
	
	cin >> n >> m;
	for(int i = 1; i <= n; ++ i)	cin >> a[i];
	build(1, 1, n);

	for(int i = 1; i + m - 1 <= n; ++ i)	cout << ask(1, i, i + m - 1) << endl;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
// 	freopen("1.in", "r", stdin);
	solve(); 
	return 0;
}

分块版本

#include  
#define LL long long 
using namespace std;

const int N = 1e6 + 10;
int a[N], n, m, block[N], ans[N], bl;

int ask(int l, int r)
{
	int p = block[l], q = block[r], res = 1e9;
	if(p == q)	for(int i = p; i <= q; ++ i)	res = min(res, a[i]);
	else
	{
		for(int i = p + 1; i <= q - 1; ++ i)	res = min(res, ans[i]);
		for(int i = l; i <= p * bl; ++ i)	res = min(res, a[i]);
		for(int i = (q - 1) * bl + 1; i <= r; ++ i)	res = min(res, a[i]);
	}
	return res;
}

void solve()
{
	cin >> n >> m;
	for(int i = 1; i <= n; ++ i)	cin >> a[i];
	bl = sqrt(n);
	memset(ans, 0x3f, sizeof ans);
	for(int i = 1; i <= n; ++ i)
	{
		block[i] = (i - 1) / bl + 1;
		ans[block[i]] = min(ans[block[i]], a[i]);
	}	
	for(int i = 1; i + m - 1 <= n; ++ i)	cout << ask(i, i + m - 1) << endl;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
//	freopen("1.in", "r", stdin);
	solve(); 
	return 0;
}