【BZOJ 4456】【UOJ #184】【ZJOI 2016】旅行者

http://www.lydsy.com/JudgeOnline/problem.php?id=4456

http://uoj.ac/problem/184

参考(抄)的晨爷的题解(代码)

对矩形进行分治。

每次对一个分治中的矩形，枚举中轴线上的点，依次做dijkstra，范围是该矩形内的点。

处理出中轴线上的点到矩形内所有点的最短路，这样，两点在该矩形内的询问就可以用$dist+dist$更新了，意义是两点经过该中轴线的最短路。

在把矩形劈成两半，把询问也分成两半，递归分治。

因为两点间的最短路一定会穿过其中一个分治矩形的中轴线，所以这么做是正确的。

时间复杂度是$O(n\sqrt{n}log^2n)$，不理解少一个$log$的复杂度的做法。

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 20003;
const int Qnum = 500003;
int in() {
	int k = 0, fh = 1; char c = getchar();
	for(; c < '0' || c > '9'; c = getchar())
		if (c == '-') fh = -1;
	for(; c >= '0' && c <= '9'; c = getchar())
		k = (k << 3) + (k << 1) + c - '0';
	return k * fh;
}

bool inq[N];
struct node {int nxt, to, w;} E[N << 2];
struct query {int x0, y0, x1, y1, id;} Q[Qnum], sta[Qnum];
int q, tot = 0, cnt = 0, point[N], n, m, ans[Qnum], dist[N];

void ins(int u, int v, int w) {E[++cnt] = (node) {point[u], v, w}; point[u] = cnt;}
int get(int x, int y) {return (x - 1) * m + y;}

struct Point {
	int id, dist;
	Point(int _id = 0, int _dist = 0) : id(_id), dist(_dist) {}
	bool operator < (const Point &A) const {
		return dist > A.dist;
	}
};
priority_queue <Point> qu;
void dijkstra(int S, int x0, int y0, int x1, int y1) {
	int u;
	for(int i = x0; i <= x1; ++i)
		for(int j = y0; j <= y1; ++j) {
			u = get(i, j);
			dist[u] = 0x7fffffff; inq[u] = false;
		}
	dist[S] = 0;
	Point x;
	qu.push(Point(S, 0));
	while (!qu.empty()) {
		x = qu.top();
		qu.pop();
		if (inq[x.id]) continue;
		inq[x.id] = true;
		for(int i = point[x.id]; i; i = E[i].nxt)
			if (x.dist + E[i].w < dist[E[i].to]) {
				dist[E[i].to] = x.dist + E[i].w;
				qu.push(Point(E[i].to, dist[E[i].to]));
			}
	}
}

void cdq(int x0, int y0, int x1, int y1, int Ql, int Qr) {
	if (Ql > Qr) return;
	if (x1 - x0 > y1 - y0) {
		int mid = (x1 + x0) >> 1, po, tmp_l, tmp_r;
		for(int i = y0; i <= y1; ++i) {
			po = get(mid, i);
			dijkstra(po, x0, y0, x1, y1);
			for(int j = Ql; j <= Qr; ++j)
				ans[Q[j].id] = min(ans[Q[j].id], dist[get(Q[j].x0, Q[j].y0)] + dist[get(Q[j].x1, Q[j].y1)]);
		}

		tmp_l = Ql - 1; tmp_r = Qr + 1;
		for(int i = Ql; i <= Qr; ++i)
			if (Q[i].x0 < mid && Q[i].x1 < mid) sta[++tmp_l] = Q[i];
			else if (Q[i].x0 > mid && Q[i].x1 > mid) sta[--tmp_r] = Q[i];
		for(int i = Ql; i <= tmp_l; ++i) Q[i] = sta[i];
		for(int i = tmp_r; i <= Qr; ++i) Q[i] = sta[i];

		cdq(x0, y0, mid - 1, y1, Ql, tmp_l);
		cdq(mid + 1, y0, x1, y1, tmp_r, Qr);
	} else {
		int mid = (y0 + y1) >> 1, po, tmp_l, tmp_r;
		for(int i = x0; i <= x1; ++i) {
			po = get(i, mid);
			dijkstra(po, x0, y0, x1, y1);
			for(int j = Ql; j <= Qr; ++j)
				ans[Q[j].id] = min(ans[Q[j].id], dist[get(Q[j].x0, Q[j].y0)] + dist[get(Q[j].x1, Q[j].y1)]);
		}

		tmp_l = Ql - 1; tmp_r = Qr + 1;
		for(int i = Ql; i <= Qr; ++i)
			if (Q[i].y0 < mid && Q[i].y1 < mid) sta[++tmp_l] = Q[i];
			else if (Q[i].y0 > mid && Q[i].y1 > mid) sta[--tmp_r] = Q[i];
		for(int i = Ql; i <= tmp_l; ++i) Q[i] = sta[i];
		for(int i = tmp_r; i <= Qr; ++i) Q[i] = sta[i];

		cdq(x0, y0, x1, mid - 1, Ql, tmp_l);
		cdq(x0, mid + 1, x1, y1, tmp_r, Qr);
	}
}

int main() {
	n = in(); m = in();
	int len, Point;
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j < m; ++j) {
			len = in();
			Point = get(i, j);
			ins(Point, Point + 1, len);
			ins(Point + 1, Point, len);
		}
	for(int i = 1; i < n; ++i)
		for(int j = 1; j <= m; ++j) {
			len = in();
			Point = get(i, j);
			ins(Point, Point + m, len);
			ins(Point + m, Point, len);
		}

	q = in();
	int x0, y0, x1, y1;
	memset(ans, 127, sizeof(int) * (q + 1));
	for(int i = 1; i <= q; ++i) {
		x0 = in(); y0 = in(); x1 = in(); y1 = in();
		if (x0 == x1 && y0 == y1) {ans[i] = 0; continue;}
		Q[++tot] = (query) {x0, y0, x1, y1, i};
	}

	memset(inq, 1, sizeof(bool) * (n * m + 3));
	cdq(1, 1, n, m, 1, tot);

	for(int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
	return 0;
}

终于AC了