YBTOJLuogu P1527[国家集训队]矩阵乘法

Posted 2021-06-12 Jayun

链接：

洛谷

题目大意：

二维区间第 \\(k\\) 小。

正文：

就是区间第 \\(k\\) 小的升维版，详见 区间第 \\(k\\) 小。

代码：

const int N = 510 + 10, M = 6e4 + 10, P = 25e4 + 10;

inline ll Read()
{
	ll x = 0, f = 1;
	char c = getchar();
	while (c != \'-\' && (c < \'0\' || c > \'9\')) c = getchar();
	if (c == \'-\') f = -f, c = getchar();
	while (c >= \'0\' && c <= \'9\') x = (x << 3) + (x << 1) + c - \'0\', c = getchar();
	return x * f;
}

int n, m, tot;
ll t[N][N], ans[M];

void modify(int x, int y, ll val) 
{
	for (int i = x; i <= n; i += i & -i) 
		for (int j = y; j <= n; j += j & -j) 
			t[i][j] += val;
} 
int Query(int x, int y) 
{
	int ans = 0;
	for (int i = x; i; i -= i & -i) 
		for (int j = y; j; j -= j & -j) 
			ans += t[i][j]; 
	return ans;
} 

int query(int Lx, int Ly, int Rx, int Ry)
{
	return Query(Rx, Ry) - Query(Lx - 1, Ry) - Query(Rx, Ly - 1) + Query(Lx - 1, Ly - 1);
}

struct node
{
	int Lx, Ly, Rx, Ry, k, id;
	node(){	}
	node(int a, int b, int c, int d, int val, int i) 
	{
		Lx = a, Ly = b, Rx = c, Ry = d, k = val, id = i;
	} 
}q[P + M], q1[P + M], q2[P + M];

void Solve (int l, int r, int L, int R)
{
	if (l > r || L > R) return ; 
	if(l == r)
	{
		for (int i = L; i <= R; i++) if(q[i].id) ans[q[i].id] = l;
		return ;
	}
	int mid = l + r >> 1;
	int n1 = 0, n2 = 0;
	for (int i = L; i <= R; i ++)
		if (q[i].id) 
		{
			ll tmp = query(q[i].Lx, q[i].Ly, q[i].Rx, q[i].Ry);
			if (q[i].k <= tmp) q1[++n1] = q[i];
			else q[i].k -= tmp, q2[++n2] = q[i];
		}
		else
			if (q[i].k <= mid) modify(q[i].Lx, q[i].Ly, 1), q1[++n1] = q[i];
			else q2[++n2] = q[i];
	for (int i = 1; i <= n1; i++) if(!q1[i].id) modify(q1[i].Lx, q1[i].Ly, -1);
	
	for (int i = 1; i <= n1; i++) q[L + i - 1] = q1[i];
	for (int i = 1; i <= n2; i++) q[L + n1 + i - 1] = q2[i];
	
	Solve(l, mid, L, L + n1 - 1), Solve(mid + 1, r, L + n1, R);
}

int main()
{
	n = Read(), m = Read();
	for (int i = 1; i <= n; i++) 
		for (int j = 1; j <= n; j++)
			q[++tot] = node(i, j, 0, 0, Read(), 0);
	for (int i = 1; i <= m; i++) 
		q[++tot].Lx = Read(), q[tot].Ly = Read(), q[tot].Rx = Read(), q[tot].Ry = Read(), 
		q[tot].k = Read(), q[tot].id = i;
	Solve(0, 1e9, 1, tot);
	for (int i = 1; i <= m; i++) printf ("%lld\\n", ans[i]);
	return 0;
}