路径规划中我只想要两条路径的c++代码

以下是一种简单的方法，使用 Dijkstra 算法来计算两条最短路径：

#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
using namespace std;

const int MAXN = 1005;
const int INF = 1e9;

struct Edge {
    int to, w;
};

vector<Edge> graph[MAXN];
int dist[MAXN];
int parent[MAXN];

void dijkstra(int s) {
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
    memset(dist, INF, sizeof(dist));
    dist[s] = 0;
    pq.push(make_pair(0, s));
    while (!pq.empty()) {
        int u = pq.top().second;
        pq.pop();
        for (int i = 0; i < graph[u].size(); i++) {
            int v = graph[u][i].to;
            int w = graph[u][i].w;
            if (dist[v] > dist[u] + w) {
                dist[v] = dist[u] + w;
                pq.push(make_pair(dist[v], v));
                parent[v] = u;
            }
        }
    }
}

int main() {
    int n, m, s, t;
    cin >> n >> m >> s >> t;
    for (int i = 0; i < m; i++) {
        int u, v, w;
        cin >> u >> v >> w;
        graph[u].push_back({v, w});
        graph[v].push_back({u, w});
    }
    dijkstra(s);
    int path1[MAXN], path2[MAXN], len1 = 0, len2 = 0;
    int cur = t;
    while (cur != s) {
        path1[len1++] = cur;
        cur = parent[cur];
    }
    path1[len1++] = s;
    reverse(path1, path1 + len1);
    dijkstra(t);
    cur = s;
    while (cur != t) {
        path2[len2++] = cur;
        cur = parent[cur];
    }
    path2[len2++] = t;
    reverse(path2, path2 + len2);
    cout << "Path 1: ";
    for (int i = 0; i < len1; i++) {
        cout << path1[i];
        if (i < len1 - 1) cout << " -> ";
    }
    cout << endl;
    cout << "Path 2: ";
    for (int i = 0; i < len2; i++) {
        cout << path2[i];
        if (i < len2 - 1) cout << " -> ";
    }
    cout << endl;
    return 0;
}

该代码首先读取输入并构建邻接表表示的图。然后运行两次 Dijkstra 算法，一次从起点到终点，一次从终点到起点。

在计算两次最短路径时，我们记录每个节点的父节点，以便后续从终点向起点回溯路径。最后，我们将两个路径打印出来。

请注意，如果两条路径中存在公共路径，则此代码将忽略该路径并选择两个独立的路径。如果需要包括公共路径，则需要在计算第二条路径时使用 Dijkstra 算法，并记录每个节点的距离和两个父节点，以便在回溯路径时合并公共路径。