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链接：https://ac.nowcoder.com/acm/contest/945/D
来源：牛客网
Description
Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.
The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1…N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.
The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).
Input
Line 1: Two space-separated integers: N and R
Lines 2…R+1: Each line contains three space-separated integers: A, B, and D that describe a road that connects intersections A and B and has length D (1 ≤ D ≤ 5000)
Output
Line 1: The length of the second shortest path between node 1 and node N
示例1
Input Sample
4 4
1 2 100
2 4 200
2 3 250
3 4 100
Output Sample
450
说明
Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)
解题思路
求次短路：从两头用dijkstra求出最短路，得到两个数组，遍历一遍所有的边，从每条边（u，v）中，找到所有满足 dis1[u]+EDGE(u,v)+dis2[v] > minDis（起点到终点的最短距离）条件下的最小值，即使解
AC Code
/*
* Copyright (c) 2019 Ng Kimbing, HNU, All rights reserved. May not be used, modified, or copied without permission.
* @Author: Ng Kimbing, HNU.
* @LastModified:2019-06-25 T 10:43:37.961 +08:00
*/
package ACMProblems.QianDaoTi;
import java.util.*;
import static ACMProblems.ACMIO.*;
/*
* 链接：https://ac.nowcoder.com/acm/contest/945/D
* 来源：牛客网
*
* ## Description
* Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.
* The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.
* The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).
* ## Input
* Line 1: Two space-separated integers: N and R
* Lines 2..R+1: Each line contains three space-separated integers: A, B, and D that describe a road that connects intersections A and B and has length D (1 ≤ D ≤ 5000)
*
* ## Output
* Line 1: The length of the second shortest path between node 1 and node N
* 示例1
* ## Input Sample
* >4 4
* 1 2 100
* 2 4 200
* 2 3 250
* 3 4 100
*
* ## Output Sample
* >450
* >
* **说明**
* Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)
*/
public class SecondShortest {
static class VertexNode {
private int id;
private int weight;
public VertexNode(int id, int weight) {
this.id = id;
this.weight = weight;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public int getWeight() {
return weight;
}
public void setWeight(int weight) {
this.weight = weight;
}
@Override
public boolean equals(Object obj) {
return this.id == ((VertexNode) obj).id;
}
}
static class Edge {
int from;
int to;
int weight;
public Edge(int from, int to, int weight) {
this.from = from;
this.to = to;
this.weight = weight;
}
@Override
public String toString() {
return "(" + from + " -> " + to + ", " + weight + ")";
}
}
private static int vertexNum;
private static ArrayList<VertexNode>[] lists;
/**
* Add one edge into the graph. The weight will be changed if the edge already exists.
*
* @param a Start
* @param b End
* @param weight The weight of the edge from a to b
*/
public static void addEdge(int a, int b, int weight) {
assert a < vertexNum && b < vertexNum;
lists[a].add(new VertexNode(b, weight));
lists[b].add(new VertexNode(a, weight));
edges[index++] = new Edge(a, b, weight);
edges[index++] = new Edge(b, a, weight);
}
private static int[] dijkstra(int a, int b) {
assert a < vertexNum && b < vertexNum;
int[] dis = new int[vertexNum];
boolean[] visited = new boolean[vertexNum];
// The heap used to find the closest vertex.
PriorityQueue<VertexNode> q = new PriorityQueue<>(vertexNum, Comparator.comparingInt(VertexNode::getWeight));
//initialize
q.add(new VertexNode(a, 0));
Arrays.fill(dis, 0x3f3f3f3f);
dis[a] = 0;
while (!q.isEmpty()) {
// Current vertex.
VertexNode v = q.poll();
int currID = v.getId();
//Mark as visited.
visited[currID] = true;
//for each neighbor
for (VertexNode dest : lists[currID]) {
//if there is a shorter path
if (!visited[dest.getId()] && dis[currID] + dest.getWeight() < dis[dest.getId()]) {
dis[dest.getId()] = dis[currID] + dest.getWeight();
q.add(new VertexNode(dest.getId(), dis[dest.getId()]));
}
}
}
return dis;
}
static Edge[] edges;
static int index = 0;
@SuppressWarnings("unchecked")
public static void main(String[] args) throws Exception {
setStream(System.in);
vertexNum = nextInt();
int edgeNum = nextInt();
edges = new Edge[edgeNum * 2];
lists = new ArrayList[vertexNum];
for (int i = 0; i < vertexNum; ++i)
lists[i] = new ArrayList<>();
for (int i = 0; i < edgeNum; ++i) {
int a = nextInt() - 1;
int b = nextInt() - 1;
int weight = nextInt();
addEdge(a, b, weight);
}
int[] dis1 = dijkstra(0, vertexNum - 1);
int[] dis2 = dijkstra(vertexNum - 1, 0);
int minDis = dis1[vertexNum - 1];
int currMin = Integer.MAX_VALUE;
for (Edge edge : edges) {
int from = edge.from;
int to = edge.to;
int weight = edge.weight;
int foo = dis1[from] + dis2[to] + weight;
if (foo > minDis)
currMin = Math.min(foo, currMin);
}
System.out.println(currMin);
}
}