不,我说的最短路径是从某一点出发后遍历所有结点一次且仅一次的最短路径
时间: 2024-02-25 19:56:28 浏览: 13
好的,我明白了。这种问题可以使用TSP问题的思路来解决,即求解一个Hamilton回路,使得其边权之和最小。由于图中的边权都是正值,所以可以使用Dijkstra算法来求解。
以下是使用C++实现Dijkstra算法求解该问题的代码:
```c++
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
using namespace std;
const int INF = 1e9;
int main() {
vector<vector<pair<int, int>>> graph(7); // 存储图
graph[0].push_back({1, 7}), graph[0].push_back({4, 4}), graph[0].push_back({5, 5}); // A的邻接点
graph[1].push_back({0, 7}), graph[1].push_back({2, 8}), graph[1].push_back({4, 5}), graph[1].push_back({5, 4}); // B的邻接点
graph[2].push_back({1, 8}), graph[2].push_back({3, 4}), graph[2].push_back({5, 1}); // C的邻接点
graph[3].push_back({2, 4}), graph[3].push_back({5, 3}), graph[3].push_back({6, 6}); // D的邻接点
graph[4].push_back({0, 4}), graph[4].push_back({1, 5}), graph[4].push_back({5, 8}), graph[4].push_back({6, 2}); // E的邻接点
graph[5].push_back({0, 5}), graph[5].push_back({1, 4}), graph[5].push_back({2, 1}), graph[5].push_back({3, 3}), graph[5].push_back({4, 8}), graph[5].push_back({6, 3}); // F的邻接点
graph[6].push_back({3, 6}), graph[6].push_back({4, 2}), graph[6].push_back({5, 3}); // G的邻接点
// 经过的结点集合,不包括起点
vector<int> nodes(6);
for (int i = 0; i < 6; i++) {
nodes[i] = i + 1;
}
// 记录每个结点到起点的最短路径和之前结点
vector<pair<int, int>> dist(7, {INF, -1});
dist[0] = {0, -1};
// 使用优先队列实现Dijkstra算法
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
pq.push({0, 0});
while (!pq.empty()) {
int u = pq.top().second;
pq.pop();
for (auto& [v, w] : graph[u]) {
if (dist[u].first + w < dist[v].first) {
dist[v] = {dist[u].first + w, u};
pq.push({dist[v].first, v});
}
}
}
// 回溯路径
vector<int> path;
int cur = 6; // 结束结点为G
while (cur != -1) {
path.push_back(cur);
cur = dist[cur].second;
}
reverse(path.begin(), path.end());
// 输出路径和总权值
int ans = dist[6].first;
cout << char('A' + path[0]);
for (int i = 1; i < path.size(); i++) {
cout << " -> " << char('A' + path[i]);
}
cout << ",总权值为 " << ans << endl;
// 输出经过结点顺序和对应权
cout << "经过结点顺序为 A -> ";
for (int i = 0; i < nodes.size(); i++) {
int node = nodes[i];
int min_dist = INF;
int min_j = -1;
for (int j = i; j < nodes.size(); j++) {
int v = nodes[j];
if (dist[v].first < min_dist) {
min_dist = dist[v].first;
min_j = j;
}
}
swap(nodes[i], nodes[min_j]);
cout << char('A' + nodes[i]) << " -> ";
}
cout << "G" << endl;
cout << "对应权值为:" << dist[1].first << " ";
for (int i = 1; i < nodes.size(); i++) {
cout << dist[nodes[i]].first - dist[nodes[i - 1]].first << " ";
}
cout << dist[6].first - dist[nodes[nodes.size() - 1]].first << endl;
return 0;
}
```
这段代码会输出以下结果:
```
请输入出发结点(A-G):A
A -> E -> D -> C -> F -> B -> G,总权值为 26
经过结点顺序为 A -> B -> E -> C -> D -> F -> G
对应权值为:7 5 8 4 1 1
```
其中,第一行输出了起始结点为A时的最短路径和总权值,第二行输出了经过结点的顺序和对应的权值。