C++ 具有节点访问代价的Dijkstra算法?
所以我一直在关注这个 所以我想补充一点,每个节点都有某种数量或价格,这是我的代码:C++ 具有节点访问代价的Dijkstra算法?,c++,dijkstra,C++,Dijkstra,所以我一直在关注这个 所以我想补充一点,每个节点都有某种数量或价格,这是我的代码: #include <iostream> #include <bits/stdc++.h> #include <vector> #define INF 0x3f3f3f3f typedef std::pair<int, int> iPair; std::unordered_map<int, int> hashTable;
#include <iostream>
#include <bits/stdc++.h>
#include <vector>
#define INF 0x3f3f3f3f
typedef std::pair<int, int> iPair;
std::unordered_map<int, int> hashTable;
class Graph
{
int V;
std::list<std::pair<int, int>> *adj;
public:
Graph(int V);
void addEdge(int u, int v, int w);
void shortestPath(int s, int end);
};
Graph::Graph(int V)
{
this->V = V;
adj = new std::list<iPair>[V];
}
void Graph::addEdge(int u, int v, int w)
{
adj[u].push_back(std::make_pair(v, w));
adj[v].push_back(std::make_pair(u, w));
}
void Graph::shortestPath(int src, int end)
{
std::priority_queue<iPair, std::vector<iPair>, std::greater<iPair>> pq;
std::vector<int> dist(V, INF);
pq.push(std::make_pair(0, src));
dist[src] = 0;
while (!pq.empty())
{
int u = pq.top().second;
pq.pop();
std::list<std::pair<int, int>>::iterator i;
for (i = adj[u].begin(); i != adj[u].end(); ++i)
{
int v = (*i).first;
int weight = (*i).second;
if (dist[v] > dist[u] + weight + hashTable[v])
{
dist[v] = dist[u] + weight + hashTable[v];
pq.push(std::make_pair(dist[v], v));
}
}
}
printf("Vertex Distance from Source\n");
for (int i = 0; i < V; ++i)
printf("%d \t\t %d\n", i, dist[i]);
}
int main()
{
int V = 9;
Graph g(V);
g.addEdge(0, 1, 4);
g.addEdge(0, 7, 8);
g.addEdge(1, 2, 8);
g.addEdge(1, 7, 11);
g.addEdge(2, 3, 7);
g.addEdge(2, 8, 2);
g.addEdge(2, 5, 4);
g.addEdge(3, 4, 9);
g.addEdge(3, 5, 14);
g.addEdge(4, 5, 10);
g.addEdge(5, 6, 2);
g.addEdge(6, 7, 1);
g.addEdge(6, 8, 6);
g.addEdge(7, 8, 7);
hashTable[2] = 30;
g.shortestPath(0, 2);
return 0;
}
这是通过哈希表输出的:
0 0
1 4
2 12
3 19
4 21
5 11
6 9
7 8
8 14
0 0
1 4
2 42
3 25
4 21
5 11
6 9
7 8
8 15
有人能解释一下为什么这不起作用吗?我对算法这个话题还很陌生,所以我试图理解它。为什么
hashTable[2]=30代码>这30是12的额外值,它为您提供了42个,但为什么id 8上的节点在哈希表之前有14个,而它却有15个成本?