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📜  从源到目标的边缘数为偶数的最短路径

📅  最后修改于: 2021-06-26 08:53:55             🧑  作者: Mango

给定无向图G ,任务是找到偶数长度的最短路径,给定1作为源节点,给定N作为目标节点。路径长度是指路径中存在的边数(而不是路径成本)。

例子:

方法:
创建一个新图( G’ )。对于初始图G中的每个节点V ,创建两个新节点V_evenV_odd

现在,对于G中的每个边( U,V ),在G’中添加两个新边, (U_even,V_odd)(U_odd,V_even) 。最后,使用Dijkstra最短路径算法找到从(source_even)节点到(destination_even)节点的最短路径。
对于输入1(上方)中给出的图, G’可以表示为:

图G'

从图G’可以看出,只有从(1_even)(5_even)的长度路径。因此,奇数长度的路径以G’分开,并且可以获得所需的最短路径。

下面是上述方法的实现:

C++
// C++ program for the above approach
#include 
using namespace std;
 
const int MAXX = 10000, INF = 1e9;
 
// Adjacency List: to represent graph
vector > >
    adj(MAXX * 10 + 3);
 
// Distance Array: to store shortest
// distance to every node
vector dist(MAXX * 10 + 3, INF);
 
// returns value which will
// represent even_x
int even(int x)
{
    return x * 10 + 2;
}
// returns value which will
// represent odd_x
int odd(int x)
{
    return x * 10 + 1;
}
 
// converting edge (a->b) to 2
// different edges i.e. (a->b)
// converts to (1). even_a -> odd_b
// (2). odd_a -> even_b
// since, graph is undirected, so we
// push them in reverse order too
// hence, 4 push_back operations are
// there.
void addEdge(int a, int b, int cost)
{
    adj[even(a)].push_back(
        { odd(b), cost });
    adj[odd(a)].push_back(
        { even(b), cost });
    adj[odd(b)].push_back(
        { even(a), cost });
    adj[even(b)].push_back(
        { odd(a), cost });
}
 
// Function calculates shortest
// distance to all nodes from
// "source" using Dijkstra
// Shortest Path Algorithm
// and returns shortest distance
// to "destination"
int dijkstra(int source,
             int destination)
{
 
    /* Priority Queue/min-heap
    to store and process
    (distance, node) */
    priority_queue,
                   vector >,
                   greater > >
        pq;
 
    // pushing source node to
    // priority queue and dist from
    // source to source is set to 0
    pq.push({ 0, even(source) });
    dist[even(source)] = 0;
 
    while (!pq.empty()) {
 
        // U is the node at top
        // of the priority queue
        // note that pq.top().first
        // refers to the Distance
        // and pq.top().second
        // will refer to the Node
        int u = pq.top().second;
        pq.pop();
 
        // exploring all neighbours
        // of node u
        for (pair p :
             adj[u]) {
 
            /* v is neighbour node of u
          and c is the cost/weight
          of edge (u, v) */
            int v = p.first;
            int c = p.second;
 
            // relaxation: checking if there
            // is a shorter path to v via u
            if (dist[u] + c
                < dist[v]) {
 
                // updating distance of v
                dist[v] = dist[u] + c;
                pq.push({ dist[v], v });
            }
        }
    }
 
    // returning shortest
    // distance to "destination"
    return dist[even(destination)];
}
 
// Driver function
int main()
{
    // n = number of Nodes,
    // m = number of Edges
    int n = 5, m = 6;
    addEdge(1, 2, 1);
    addEdge(2, 3, 2);
    addEdge(2, 5, 15);
    addEdge(3, 5, 1);
    addEdge(3, 4, 4);
    addEdge(5, 4, 3);
 
    int source = 1;
    int destination = n;
    int ans = dijkstra(source, destination);
 
    // if ans is INF: There is no
    // even length path from source
    // to destination else path
    // exists and we print the
    // shortest distance
    if (ans == INF)
        cout << "-1"
             << "\n";
    else
        cout << ans << "\n";
 
    return 0;
}


Java
// Java program for the above approach
import java.util.ArrayList;
import java.util.Arrays;
import java.util.PriorityQueue;
 
class GFG{
 
static class Pair implements Comparable
{
    int first, second;
 
    public Pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
 
    @Override
    public int compareTo(GFG.Pair o)
    {
        if (this.first == o.first)
        {
            return this.second - o.second;
        }
        return this.first - o.first;
    }
}
 
static final int MAXX = 10000, INF = (int)1e9;
 
// Adjacency List: to represent graph
@SuppressWarnings("unchecked")
static ArrayList[] adj = new ArrayList[MAXX * 10 + 3];
 
// Distance Array: to store shortest
// distance to every node
static int[] dist = new int[MAXX * 10 + 3];
 
// Returns value which will
// represent even_x
static int even(int x)
{
    return x * 10 + 2;
}
 
// Returns value which will
// represent odd_x
static int odd(int x)
{
    return x * 10 + 1;
}
 
// Converting edge (a->b) to 2
// different edges i.e. (a->b)
// converts to (1). even_a -> odd_b
// (2). odd_a -> even_b
// since, graph is undirected, so we
// push them in reverse order too
// hence, 4 push_back operations are
// there.
static void addEdge(int a, int b, int cost)
{
    adj[even(a)].add(new Pair(odd(b), cost));
    adj[odd(a)].add(new Pair(even(b), cost));
    adj[odd(b)].add(new Pair(even(a), cost));
    adj[even(b)].add(new Pair(odd(a), cost));
}
 
// Function calculates shortest
// distance to all nodes from
// "source" using Dijkstra
// Shortest Path Algorithm
// and returns shortest distance
// to "destination"
static int dijkstra(int source, int destination)
{
     
    // Priority Queue/min-heap to store
    // and process (distance, node)
    PriorityQueue pq = new PriorityQueue<>();
 
    // Pushing source node to
    // priority queue and dist from
    // source to source is set to 0
    pq.add(new Pair(0, even(source)));
    dist[even(source)] = 0;
 
    while (!pq.isEmpty())
    {
         
        // U is the node at top
        // of the priority queue
        // note that pq.top().first
        // refers to the Distance
        // and pq.top().second
        // will refer to the Node
        int u = pq.poll().second;
 
        // Exploring all neighbours
        // of node u
        for(Pair p : adj[u])
        {
             
            // v is neighbour node of u and
            // c is the cost/weight of edge (u, v)
            int v = p.first;
            int c = p.second;
 
            // Relaxation: checking if there
            // is a shorter path to v via u
            if (dist[u] + c < dist[v])
            {
                 
                // Updating distance of v
                dist[v] = dist[u] + c;
                pq.add(new Pair(dist[v], v));
            }
        }
    }
     
    // Returning shortest
    // distance to "destination"
    return dist[even(destination)];
}
 
// Driver code
public static void main(String[] args)
{
    for(int i = 0; i < MAXX * 10 + 3; i++)
    {
        adj[i] = new ArrayList();
    }
 
    Arrays.fill(dist, INF);
 
    // n = number of Nodes,
    // m = number of Edges
    int n = 5, m = 6;
    addEdge(1, 2, 1);
    addEdge(2, 3, 2);
    addEdge(2, 5, 15);
    addEdge(3, 5, 1);
    addEdge(3, 4, 4);
    addEdge(5, 4, 3);
 
    int source = 1;
    int destination = n;
    int ans = dijkstra(source, destination);
 
    // If ans is INF: There is no
    // even length path from source
    // to destination else path
    // exists and we print the
    // shortest distance
    if (ans == INF)
        System.out.println("-1");
    else
        System.out.println(ans);
}
}
 
// This code is contributed by sanjeev2552


输出:
10

时间复杂度: (E * log(V))

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