洛谷 Problem P1629 - 邮递员送信
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;
}
}