【JZOJ6404】【NOIP2019模拟11.04】B

题目大意

给出一个(n*m)的矩阵和一个整数(k)。设(f(i,j))表示以((i,j))为左上角，边长为(k)的正方形内权值的种类数。
你要求(f(i,j))的总和和最大值。

(n,mleq 3000,a_{i,j}leq100000)

Solution

用扫描线维护(k)列中(n-k+1)个正方形的答案，考虑向右移动一列的影响。
那么有一列要加，一列要删。
我们对于每个权值，维护一个bitset，记录这个权值在(k)列中的哪几行出现过。对于要删的数，我们查询前驱后继，就能知道删去它会使哪些行答案(-1)。显然这些行是连续的，加入一个数同理。那么用差分维护区间修改，就行了。

复杂度(O(frac{nm^2}{omega}))。

Code

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 3007, M = 100007;
inline int read() {
	int x = 0, f = 0;
	char c = getchar();
	for (; c < '0' || c > '9'; c = getchar()) if (c == '-') f = 1;
	for (; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ '0');
	return f ? -x : x;
}

int n, m, k, cnt, ans[N], a[N][N], buc[M], c[N];
bool g[N][N];
int ans1; long long ans2;

struct bitset {
	unsigned _[95];
	void set(int po) { _[po >> 5] |= (1 << (po & 31)); }
	void del(int po) { _[po >> 5] -= (1 << (po & 31)); }
	int get(int po) { return (_[po >> 5] >> (po & 31)) & 1; }
	int pre(int po) {
		int a = po >> 5, b = po & 31;
		for (int i = b - 1; i >= 0; --i) if (_[a] & (1 << i)) return po - b + i;
		for (int i = a - 1; i >= 0; --i) if (_[i]) for (int j = 31; j >= 0; --j) if (_[i] & (1 << j)) return (i << 5) + j;
	}
	int nxt(int po) {
		int a = po >> 5, b = po & 31;
		for (int i = b + 1; i < 32; ++i) if (_[a] & (1 << i)) return po + i - b;
		for (int i = a + 1; i < 94; ++i) if (_[i]) for (int j = 0; j < 32; ++j) if (_[i] & (1 << j)) return (i << 5) + j;
	}
} b[M];

void upd(int res) { ans1 = max(ans1, res), ans2 += res; }
void plus(int l, int r, int v) { if (l <= r) c[l] += v, c[r + 1] -= v; }
void updans() {
	for (int i = 1; i <= n; ++i) ans[i] += (c[i] += c[i - 1]);
	for (int i = 1; i <= n; ++i) c[i] = 0;
}

int main() {
	freopen("b.in", "r", stdin);
	//freopen("b.out", "w", stdout);
	n = read(), m = read(), k = read();
	for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) a[i][j] = read();
	cnt = 0;
	for (int i = 1; i <= k; ++i) for (int j = 1; j <= k; ++j) cnt += (buc[a[i][j]] == 0), ++buc[a[i][j]];
	ans[1] = cnt, upd(ans[1]);
	for (int i = 1; i <= n - k; ++i) {
		for (int j = 1; j <= k; ++j) --buc[a[i][j]], cnt -= (buc[a[i][j]] == 0);
		for (int j = 1; j <= k; ++j) cnt += (buc[a[i + k][j]] == 0), ++buc[a[i + k][j]];
		ans[i + 1] = cnt, upd(ans[i + 1]);
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) buc[a[i][j]] = 0x3f3f3f3f;
		for (int j = m; j >= 1; --j) g[i][j] = (buc[a[i][j]] - j) <= k, buc[a[i][j]] = j;
	}
	for (int i = 1; i <= n; ++i) for (int j = 1; j <= k; ++j) b[a[i][j]].set(i);
	for (int i = 1; i <= 100000; ++i) b[i].set(0), b[i].set(n + 1);
	for (int j = 1; j <= m - k; ++j) {
		for (int i = 1; i <= n; ++i)
			if (!g[i][j]) {
				int a1 = b[a[i][j]].pre(i) + 1, a2 = b[a[i][j]].nxt(i) - 1;
				plus(max(a1, i - k + 1), min(a2 - k + 1, i), -1);
				b[a[i][j]].del(i);
			}
		for (int i = 1; i <= n; ++i)
			if (!b[a[i][j + k]].get(i)) {
				int a1 = b[a[i][j + k]].pre(i) + 1, a2 = b[a[i][j + k]].nxt(i) - 1;
				plus(max(a1, i - k + 1), min(a2 - k + 1, i), 1);
				b[a[i][j + k]].set(i);
			}
		updans();
		for (int i = 1; i <= n; ++i) upd(ans[i]);
	}
	printf("%d %lld
", ans1, ans2);
	return 0;
}