Loj#2090. 「ZJOI2016」旅行者（网格图分治）

https://loj.ac/problem/2090

题解：

考虑每次取矩形较大那一维的中线，跑dij求出每个点到中线上的点的最短距离。

对于每个询问，都枚举中线上每个点为中转点问一下，按照在左边还是右边分治下去，写起来类似整体二分。

可以分析出来矩形面积*中线长度极限是当(n=m)时取(nm imes sqrt{nm})，再乘上堆优化dij的复杂度即是极限复杂度。

每个询问的复杂度是(log*min(n,m))。

Code：

#include<bits/stdc++.h>

#define fo(i, x, y)	for(int i = x, _b = y; i <= _b; i ++)

#define ff(i, x, y) for(int i = x, _b = y; i <  _b; i ++)

#define fd(i, x, y) for(int i = x, _b = y; i >= _b; i --)

#define ll long long

#define pp printf

#define hh pp("
")

using namespace std;

const int N = 20005;

int n, m;

#define V vector<int>
#define re resize
#define pb push_back
#define si size()
V a[N], b[N];

int Q;

struct P {
	int x1, y1, x2, y2;
} q[100005];

int ans[100005];

vector<int> f[N], us[N];

const int inf = 1e9;

struct nod {
	int z, x, y;
};

bool operator < (nod a, nod b) { return a.z > b.z;}

namespace sub {
priority_queue<nod> q;
void gx(int z, int x, int y) {
	if(z < f[x][y]) {
		f[x][y] = z;
		q.push((nod) {z, x, y});
	}
}
void dp(int x1, int x2, int y1, int y2, int x, int y) {
	fo(i, x1, x2) fo(j, y1, y2) f[i][j] = inf, us[i][j] = 0;
	f[x][y] = 0; q.push((nod) {0, x, y});
	while(q.size()) {
		nod c;
		while(q.size()) {
			c = q.top(), q.pop();
			if(us[c.x][c.y]) continue;
			break;
		}
		if(us[c.x][c.y]) break;
		us[c.x][c.y] = 1;
		if(c.x > x1) gx(c.z + b[c.x - 1][c.y], c.x - 1, c.y);
		if(c.x < x2) gx(c.z + b[c.x][c.y], c.x + 1, c.y);	
		if(c.y > y1) gx(c.z + a[c.x][c.y - 1], c.x, c.y - 1);
		if(c.y < y2) gx(c.z + a[c.x][c.y], c.x, c.y + 1);
	}
}
}


void dg(int x1, int x2, int y1, int y2, V w) {
	if(x1 > x2 || y1 > y2) return;
	if(w.empty()) return;
	if(x1 == x2 && y1 == y2) {
		ff(i, 0, w.si) ans[w[i]] = 0;
		return;
	}
	V t, wl, wr;
	if(x2 - x1 >= y2 - y1) {
		int mid = x1 + x2 >> 1;
		ff(i, 0, w.si) {
			int x = w[i];
			if(q[x].x1 > q[x].x2) {
				swap(q[x].x1, q[x].x2);
				swap(q[x].y1, q[x].y2);
			}
			if(q[x].x1 <= mid && q[x].x2 >= mid) {
				t.pb(x);
			} else
			if(q[x].x2 <= mid)
				wl.pb(x); else wr.pb(x);
		}
		fo(y, y1, y2) {
			sub :: dp(x1, x2, y1, y2, mid, y);
			ff(_i, 0, w.si) {
				int i = w[_i];
				ans[i] = min(ans[i], f[q[i].x1][q[i].y1] + f[q[i].x2][q[i].y2]);
			}
		}
		dg(x1, mid, y1, y2, wl);
		dg(mid + 1, x2, y1, y2, wr);
	} else {
		int mid = y1 + y2 >> 1;
		ff(i, 0, w.si) {
			int x = w[i];
			if(q[x].y1 > q[x].y2) {
				swap(q[x].x1, q[x].x2);
				swap(q[x].y1, q[x].y2);
			}
			if(q[x].y1 <= mid && q[x].y2 >= mid) {
				t.pb(x);
			} else
			if(q[x].y2 <= mid)
				wl.pb(x); else wr.pb(x);
		}
		fo(x, x1, x2) {
			sub :: dp(x1, x2, y1, y2, x, mid);
			ff(_i, 0, w.si) {
				int i = w[_i];
				ans[i] = min(ans[i], f[q[i].x1][q[i].y1] + f[q[i].x2][q[i].y2]);
			}
		}
		dg(x1, x2, y1, mid, wl);
		dg(x1, x2, mid + 1, y2, wr);
	}
}

int main() {
	scanf("%d %d", &n, &m);
	fo(i, 1, n) {
		a[i].re(m);
		fo(j, 1, m - 1) scanf("%d", &a[i][j]);
	}
	fo(i, 1, n - 1) {
		b[i].re(m + 1);
		fo(j, 1, m) scanf("%d", &b[i][j]);
	}
	scanf("%d", &Q);
	V w;
	fo(i, 1, Q) {
		scanf("%d %d %d %d", &q[i].x1, &q[i].y1, &q[i].x2, &q[i].y2);
		w.pb(i);
		ans[i] = inf;
	}
	fo(i, 1, n) f[i].re(m + 1), us[i].re(m + 1);
	dg(1, n, 1, m, w);
	fo(i, 1, Q) pp("%d
", ans[i]);
}