题目大意：给定平面上的 $n$ 个点，定义 $(x1,y1)$ 到 $(x2,y2)$ 的费用为 $\min(|x1-x2|,|y1-y2|)$，求从 $1$ 号点走到 $n$ 号点的最小费用。$2\le n\le2\times10^5$。

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAXN = 2e5 + 10;
const int MAXM = 4e5 + 10;

struct edge {
	ll v, w, nxt;
} e[MAXM << 2];

int head[MAXN], tot;

inline 
void add(int u, int v, ll w) {
	e[++tot] = { v, w, head[u] }, head[u] = tot;
}

struct node {
	ll u, dis;
	bool operator < (const node &rhs) const { return dis > rhs.dis; }
};

priority_queue<node> q;

bool vis[MAXN];

ll dis[MAXN];

inline 
ll dijkstra(int s, int t) {
	memset(dis, 0x3f, sizeof dis);
	q.push({ s, 0 }), dis[s] = 0, vis[s] = 1;
	while (!q.empty()) {
		int u = q.top().u; q.pop();
		for (int i = head[u], v; i; i = e[i].nxt) {
			v = e[i].v;
			if (!vis[v] && dis[v] > dis[u] + e[i].w) {
				dis[v] = dis[u] + e[i].w, q.push({ v, dis[v] });
			}
		}
	}
	return dis[t];
}

struct pos {
	int id, x, y;
} a[MAXN];

bool cmpx(pos p, pos q) {
	return p.x == q.x ? p.y < q.y : p.x < q.x;
}

bool cmpy(pos p, pos q) {
	return p.y == q.y ? p.x < q.x : p.y < q.y;
}

int n;

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d%d", &a[i].x, &a[i].y), a[i].id = i;
	sort(a + 1, a + n + 1, cmpx);
	for (int i = 1, w; i < n; i++) {
		w = min(abs(a[i].x - a[i + 1].x), abs(a[i].y - a[i + 1].y));
		add(a[i].id, a[i + 1].id, w);
	}
	sort(a + 1, a + n + 1, cmpy);
	for (int i = 1, w; i < n; i++) {
		w = min(abs(a[i].x - a[i + 1].x), abs(a[i].y - a[i + 1].y));
		add(a[i].id, a[i + 1].id, w);
	}
	printf("%lld", dijkstra(1, n));
} 

样例都没过……
输入样例：

5
2 2
1 1
4 5
7 1
6 7

输出样例：

2