洛谷 Problem P1629 - 邮递员送信
2021/7/28 23:10:56
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洛谷 Problem P1629 - 邮递员送信
原题地址
题目类型:最短路径、Dijkstra
题意:
在一个有向图中,给定一个起点,对于其余的每个点进行从起点走到该点再返回起点的操作,求最终经过的总路程。
分析:
首先可以想到用 Dijkstra 算法求出从起点到各个点的最短距离,但是到达每个点后还要返回,这个路径应该也是最
短的,但因为是有向图我们无法直接求得。但是可以重新反向建图,即将原来的边反向连,然后再用 Dijkstra 算法来求最短路。此时求得的最短路就是所有点走向起点的最短路径。
代码
static int[] head1; static int[] head2; static Edge[] edge1; static Edge[] edge2; static int idx1; static int idx2; public static void solve() throws IOException { int n = nextInt(); int m = nextInt(); init(n, m); for (int i = 0; i < m; i++) { int from = nextInt(); int to = nextInt(); int dis = nextInt(); addEdge1(new Edge(from, to, dis)); addEdge2(new Edge(to, from, dis)); } int[] dis1 = dijkstra1(1, n, m); int[] dis2 = dijkstra2(1, n, m); long ans = 0; for (int i = 2; i <= n; i++) ans += dis1[i] + dis2[i]; pw.println(ans); } private static int[] dijkstra1(int start, int n, int m) { // TODO Auto-generated method stub int[] dis = new int[n + 1]; boolean[] vis = new boolean[n + 1]; Arrays.fill(dis, INF); dis[start] = 0; PriorityQueue<Vertex> q = new PriorityQueue<>(new Comparator<Vertex>() { @Override public int compare(Main.Vertex o1, Main.Vertex o2) { // TODO Auto-generated method stub return o1.dis - o2.dis; } }); q.add(new Vertex(start, 0)); while (!q.isEmpty()) { Vertex u = q.poll(); if (vis[u.ver]) continue ; vis[u.ver] = true; for (int i = head1[u.ver]; i != -1; i = edge1[i].next) { Edge e = edge1[i]; if (dis[e.to] > dis[u.ver] + e.dis) { dis[e.to] = dis[u.ver] + e.dis; q.add(new Vertex(e.to, dis[e.to])); } } } return dis; } private static int[] dijkstra2(int start, int n, int m) { int[] dis = new int[n + 1]; boolean[] vis = new boolean[n + 1]; Arrays.fill(dis, INF); dis[start] = 0; PriorityQueue<Vertex> q = new PriorityQueue<>(new Comparator<Vertex>() { @Override public int compare(Main.Vertex o1, Main.Vertex o2) { return o1.dis - o2.dis; } }); q.add(new Vertex(start, 0)); while (!q.isEmpty()) { Vertex u = q.poll(); if (vis[u.ver]) continue ; vis[u.ver] = true; for (int i = head2[u.ver]; i != -1; i = edge2[i].next) { Edge e = edge2[i]; if (dis[e.to] > dis[u.ver] + e.dis) { dis[e.to] = dis[u.ver] + e.dis; q.add(new Vertex(e.to, dis[e.to])); } } } return dis; } private static void addEdge1(Edge e) { edge1[idx1] = e; e.next = head1[e.from]; head1[e.from] = idx1++; } private static void addEdge2(Edge e) { edge2[idx2] = e; e.next = head2[e.from]; head2[e.from] = idx2++; } private static void init(int n, int m) { head1 = new int[n + 1]; head2 = new int[n + 1]; edge1 = new Edge[m]; edge2 = new Edge[m]; Arrays.fill(head1, -1); Arrays.fill(head2, -1); idx1 = 0; idx2 = 0; } /******************************************************************************************/ // Fast I/O static class Edge { int from; int to; int dis; int next; public Edge(int from, int to, int dis) { this.from = from; this.to = to; this.dis = dis; } } static class Vertex { int ver; int dis; public Vertex(int ver, int dis) { this.ver = ver; this.dis = dis; } }
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