bzoj4152 [AMPPZ2014]The Captain

Posted 2020-10-26 aziint

Description

给定平面上的 \(n\) 个点，定义 \((x_1,y_1)\) 到 \((x_2,y_2)\) 的费用为 \(\min(|x_1-x_2|,|y_1-y_2|)\) ，求从 \(1\) 号点走到 \(n\) 号点的最小费用。

Input

第一行包含一个正整数 \(n(2\le n\le 200000)\) ，表示点数。

接下来 \(n\) 行，每行包含两个整数 \(x[i],y[i] (0\le x[i],y[i]\le 10^9)\) ，依次表示每个点的坐标。

Output

一个整数，即最小费用。

Sample

Sample Input

5
2 2
1 1
4 5
7 1
6 7

Sample Output

2

Solution

这傻逼题吧...这种题都出烂了...

写这道题的原因是听说这道题卡 \(\mathrm{spfa}\) 。

#include<bits/stdc++.h>
using namespace std;

#define N 200001
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define ll long long
#define pli pair<ll, int>

inline int read() {
    int x = 0, flag = 1; char ch = getchar(); while (!isdigit(ch)) { if (!(ch ^ '-')) flag = -1; ch = getchar(); }
    while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - '0', ch = getchar(); return x * flag;
}

struct point { int x, y, id; }a[N];
bool cmpx(const point& p1, const point& p2) { return p1.x < p2.x; }
bool cmpy(const point& p1, const point& p2) { return p1.y < p2.y; }
int n, head[N], tot;
ll dis[N];
bool vis[N];
priority_queue<pli, vector<pli>, greater<pli> > q;
struct edge{ int v, next; ll w; }e[N << 2];
inline void insert(int u, int v, ll w) { e[++tot].v = v, e[tot].w = w, e[tot].next = head[u], head[u] = tot; }
#define add(u, v, w) insert(u, v, w), insert(v, u, w)

void dijkstra() {
    rep(i, 2, n) dis[i] = (1ll << 62ll);
    q.push(make_pair(0, 1));
    while (!q.empty()) {
        int u = q.top().second; q.pop();
        if (vis[u]) continue; vis[u] = 1;
        for (int i = head[u], v; i; i = e[i].next) if (dis[v = e[i].v] > dis[u] + e[i].w)
            dis[v] = dis[u] + e[i].w, q.push(make_pair(dis[v], v));
    }
}

int main() {
    n = read(); rep(i, 1, n) a[i].x = read(), a[i].y = read(), a[i].id = i;
    sort(a + 1, a + 1 + n, cmpx); rep(i, 2, n) add(a[i].id, a[i - 1].id, a[i].x - a[i - 1].x);
    sort(a + 1, a + 1 + n, cmpy); rep(i, 2, n) add(a[i].id, a[i - 1].id, a[i].y - a[i - 1].y);
    dijkstra(); printf("%lld", dis[n]);
    return 0;
}