给定一个有N个节点和E个边的有向图，一个源S和一个目标D节点。任务是找到从S到D的边的XOR总和最小的路径。如果没有从S到D的路径，则打印-1 。
例子：
Input: N = 3, E = 3, Edges = {{{1, 2}, 5}, {{1, 3}, 9}, {{2, 3}, 1}}, S = 1, and D = 3
Output: 4
The path with smallest XOR of edges weight will be 1->2->3
with XOR sum as 5^1 = 4.
Input: N = 3, E = 3, Edges = {{{3, 2}, 5}, {{3, 3}, 9}, {{3, 3}, 1}}, S = 1, and D = 3
Output: -1
方法：想法是使用Dijkstra的最短路径算法，并稍有变化。下面是解决问题的分步方法：
- 基本案例：如果源节点等于目标节点，则返回0 。
- 使用源节点及其权重为0和访问数组初始化优先级队列。
- 当优先级队列不为空时：
- 从优先级队列中弹出最上面的元素。我们称它为当前节点。
- 在被访问数组的帮助下检查当前节点是否已被访问，如果是，则继续。
- 如果当前节点是目标节点，则返回当前节点与源节点的XOR总和距离。
- 迭代与当前节点相邻的所有节点，并推入优先级队列，并将它们的距离作为XOR与当前距离和边缘权重之和。
- 否则，就没有从源到目的地的路径。因此，返回-1
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the smallest
// xor sum of edges
double minXorSumOfEdges(
int s, int d,
vector > > gr)
{
// If the source is equal
// to the destination
if (s == d)
return 0;
// Initialise the priority queue
set > pq;
pq.insert({ 0, s });
// Visited array
bool v[gr.size()] = { 0 };
// While the priority-queue
// is not empty
while (pq.size()) {
// Current node
int curr = pq.begin()->second;
// Current xor sum of distance
int dist = pq.begin()->first;
// Popping the top-most element
pq.erase(pq.begin());
// If already visited continue
if (v[curr])
continue;
// Marking the node as visited
v[curr] = 1;
// If it is a destination node
if (curr == d)
return dist;
// Traversing the current node
for (auto it : gr[curr])
pq.insert({ dist ^ it.second,
it.first });
}
// If no path exists
return -1;
}
// Driver code
int main()
{
int n = 3;
// Graph as adjacency matrix
vector > >
gr(n + 1);
// Input edges
gr[1].push_back({ 3, 9 });
gr[2].push_back({ 3, 1 });
gr[1].push_back({ 2, 5 });
// Source and destination
int s = 1, d = 3;
cout << minXorSumOfEdges(s, d, gr);
return 0;
} Java
// Java implementation of the approach
import java.util.PriorityQueue;
import java.util.ArrayList;
class Pair implements Comparable
{
int first, second;
public Pair(int first, int second)
{
this.first = first;
this.second = second;
}
@Override
public int compareTo(Pair p)
{
if (this.first == p.first)
{
return this.second - p.second;
}
return this.first - p.first;
}
}
class GFG{
// Function to return the smallest
// xor sum of edges
static int minXorSumOfEdges(int s, int d,
ArrayList> gr)
{
// If the source is equal
// to the destination
if (s == d)
return 0;
// Initialise the priority queue
PriorityQueue pq = new PriorityQueue<>();
pq.add(new Pair(0, s));
// Visited array
boolean[] v = new boolean[gr.size()];
// While the priority-queue
// is not empty
while (!pq.isEmpty())
{
// Iterator itr = pq.iterator();
// Current node
Pair p = pq.poll();
int curr = p.second;
// Current xor sum of distance
int dist = p.first;
// If already visited continue
if (v[curr])
continue;
// Marking the node as visited
v[curr] = true;
// If it is a destination node
if (curr == d)
return dist;
// Traversing the current node
for(Pair it : gr.get(curr))
pq.add(new Pair(dist ^ it.second, it.first));
}
// If no path exists
return -1;
}
// Driver code
public static void main(String[] args)
{
int n = 3;
// Graph as adjacency matrix
ArrayList> gr = new ArrayList<>();
for(int i = 0; i < n + 1; i++)
{
gr.add(new ArrayList());
}
// Input edges
gr.get(1).add(new Pair(3, 9));
gr.get(2).add(new Pair(3, 1));
gr.get(1).add(new Pair(2, 5));
// Source and destination
int s = 1, d = 3;
System.out.println(minXorSumOfEdges(s, d, gr));
}
}
// This code is contributed by sanjeev2552 Python3
# Python3 implementation of the approach
from collections import deque
# Function to return the smallest
# xor sum of edges
def minXorSumOfEdges(s, d, gr):
# If the source is equal
# to the destination
if (s == d):
return 0
# Initialise the priority queue
pq = []
pq.append((0, s))
# Visited array
v = [0] * len(gr)
# While the priority-queue
# is not empty
while (len(pq) > 0):
pq = sorted(pq)
# Current node
curr = pq[0][1]
# Current xor sum of distance
dist = pq[0][0]
# Popping the top-most element
del pq[0]
# If already visited continue
if (v[curr]):
continue
# Marking the node as visited
v[curr] = 1
# If it is a destination node
if (curr == d):
return dist
# Traversing the current node
for it in gr[curr]:
pq.append((dist ^ it[1],
it[0]))
# If no path exists
return -1
# Driver code
if __name__ == '__main__':
n = 3
# Graph as adjacency matrix
gr = [[] for i in range(n + 1)]
# Input edges
gr[1].append([ 3, 9 ])
gr[2].append([ 3, 1 ])
gr[1].append([ 2, 5 ])
# Source and destination
s = 1
d = 3
print(minXorSumOfEdges(s, d, gr))
# This code is contributed by mohit kumar 29C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the smallest
// xor sum of edges
static int minXorSumOfEdges(int s, int d, List>> gr)
{
// If the source is equal
// to the destination
if (s == d)
return 0;
// Initialise the priority queue
List> pq = new List>();
pq.Add(new Tuple(0, s));
// Visited array
int[] v = new int[gr.Count];
// While the priority-queue
// is not empty
while (pq.Count > 0)
{
pq.Sort();
// Current node
int curr = pq[0].Item2;
// Current xor sum of distance
int dist = pq[0].Item1;
// Popping the top-most element
pq.RemoveAt(0);
// If already visited continue
if(v[curr] != 0)
continue;
// Marking the node as visited
v[curr] = 1;
// If it is a destination node
if (curr == d)
return dist;
// Traversing the current node
foreach(Tuple it in gr[curr])
{
pq.Add(new Tuple(dist ^ it.Item2, it.Item1));
}
}
// If no path exists
return -1;
}
// Driver code
static void Main()
{
int n = 3;
// Graph as adjacency matrix
List>> gr = new List>>();
for(int i = 0; i < n + 1; i++)
{
gr.Add(new List>());
}
// Input edges
gr[1].Add(new Tuple(3, 9));
gr[2].Add(new Tuple(3, 1));
gr[1].Add(new Tuple(2, 5));
// Source and destination
int s = 1;
int d = 3;
Console.WriteLine(minXorSumOfEdges(s, d, gr));
}
}
// This codee is contributed by divyesh072019. 输出：
4时间复杂度： O((E + V)logV)