BZOJ 4152: [AMPPZ2014]The Captain( 最短路 )

先按x排序, 然后只有相邻节点的边才有用, 我们连起来, 再按y排序做相同操作...然后就dijkstra

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#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<queue>#include<cmath> #define rep(i, n) for(int i = 0; i < n; i++)#define clr(x, c) memset(x, c, sizeof(x)) using namespace std; const int inf = 1000000009, maxn = 200009; struct edge { int to, w; edge*next;} E[maxn << 2], *pt = E, *head[maxn]; struct node { int x, d; bool operator < (const node&o) const { return d > o.d; }}; struct P { int x, y; inline void Read() { scanf("%d%d", &x, &y); }} A[maxn]; inline void add(int u, int v, int w) { pt->to = v, pt->w = w; pt->next = head[u]; head[u] = pt++;}#define add_edge(u, v, w) add(u, v, w), add(v, u, w) bool cmpX(const int i, const int j) { return A[i].x < A[j].x;} bool cmpY(const int i, const int j) { return A[i].y < A[j].y;} int d[maxn], n, X[maxn], Y[maxn];priority_queue<node> Q; void dijkstra() { rep(i, n) d[i] = inf; d[0] = 0, Q.push( (node) {0, 0} ); while(!Q.empty()) { node t = Q.top(); Q.pop(); if(d[t.x] != t.d) continue; for(edge*e = head[t.x]; e; e = e->next) if(d[e->to] > d[t.x] + e->w) { d[e->to] = d[t.x] + e->w; Q.push( (node) {e->to, d[e->to]} ); } }} int main() { freopen("test.in", "r", stdin); cin >> n; rep(i, n) { A[i].Read(); X[i] = Y[i] = i; } sort(X, X + n, cmpX); rep(i, n - 1) { P*a = A + X[i], *b = A + X[i + 1]; if(b->x - a->x <= abs(a->y - b->y)) add_edge(X[i], X[i + 1], b->x - a->x); } sort(Y, Y + n, cmpY); rep(i, n - 1) { P*a = A + Y[i], *b = A + Y[i + 1]; if(b->y - a->y <= abs(a->x - b->x)) add_edge(Y[i], Y[i + 1], b->y - a->y); } dijkstra(); printf("%d\n", d[n - 1]); return 0;}------------------------------------------------------------------------

4152: [AMPPZ2014]The Captain

Time Limit: 20 Sec Memory Limit: 256 MB
Submit: 272 Solved: 104
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Description

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。

Input

第一行包含一个正整数n(2<=n<=200000)，表示点数。接下来n行，每行包含两个整数x[i],y[i](0<=x[i],y[i]<=10^9)，依次表示每个点的坐标。

Output

一个整数，即最小费用。

Sample Input

5
2 2
1 1
4 5
7 1
6 7

Sample Output

2

HINT

Source

鸣谢Claris上传