bzoj4165 矩阵 堆维护多路归并

题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=4165

题解

大概多路归并是最很重要的知识点了吧，近几年考察也挺多的（虽然都是作为签到题的）。

看到题目要求第 \(K\) 小矩阵，基本上可以想到用堆维护的 \(K\) 路归并。

然后我们考虑每一路是以 \((x_2, y_2)\) 为右下角的矩形的权值。那么初始的矩形的左上角应该是 \((x_2 - Mina, y_2 - Minb)\)。

于是我们用一个堆来维护这样的每一路。扩展每一路的话，因为把矩形的左边界或者上边界扩展一个单位以后权值肯定单调升，所以可以直接扩展一下左边界和上边界这两个矩形。但是这样可能会有重复的，所以拿一个 map 来判重一下。

---

这样的时间复杂度是 \(O(k\log nm)\)。

#include<bits/stdc++.h>
#include<tr1/unordered_set>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {
	int f = 0, c;
	while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
	x = c & 15;
	while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
	f ? x = -x : 0;
}

const int N = 1000 + 7;

int n, m, mina, minb, k;
int a[N][N];
ll s[N][N];

inline ll gsum(int x1, int y1, int x2, int y2) { return s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]; }

struct Matrix {
	int x1, y1, x2, y2;
	inline Matrix() {}
	inline Matrix(const int &x1, const int &y1, const int &x2, const int &y2) : x1(x1), y1(y1), x2(x2), y2(y2) {}
	inline bool operator < (const Matrix &b) const { return gsum(x1, y1, x2, y2) > gsum(b.x1, b.y1, b.x2, b.y2); }
};
std::priority_queue<Matrix> q;

std::tr1::unordered_set<ll> mp;
inline void set_mp(int x1, int y1, int x2, int y2) {
	ll v = (((((ll)x1 * m) + y1) * n + x2) * m + y2);
	mp.insert(v);
}
inline bool get_mp(int x1, int y1, int x2, int y2) {
	ll v = (((((ll)x1 * m) + y1) * n + x2) * m + y2);
	return mp.count(v);
}

inline void work() {
	for (int i = mina; i <= n; ++i)
		for (int j = minb; j <= m; ++j) q.push(Matrix(i - mina + 1, j - minb + 1, i, j)), set_mp(i - mina + 1, j - minb + 1, i, j);
	while (k--) {
		Matrix t = q.top();
		q.pop();
		if (!k) return (void)printf("%lld\n", gsum(t.x1, t.y1, t.x2, t.y2));
		if(t.x1 > 1 && !get_mp(t.x1 - 1, t.y1, t.x2, t.y2)) set_mp(t.x1 - 1, t.y1, t.x2, t.y2), q.push(Matrix(t.x1 - 1, t.y1, t.x2, t.y2));
		if(t.y1 > 1 && !get_mp(t.x1, t.y1 - 1, t.x2, t.y2)) set_mp(t.x1, t.y1 - 1, t.x2, t.y2), q.push(Matrix(t.x1, t.y1 - 1, t.x2, t.y2));
	}
}

inline void init() {
	read(n), read(m), read(mina), read(minb), read(k);
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j) read(a[i][j]), s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
}

int main() {
#ifdef hzhkk
	freopen("hkk.in", "r", stdin);
#endif
	init();
	work();
	fclose(stdin), fclose(stdout);
	return 0;
}