diff --git a/14lanqiao/test9-1.cpp b/14lanqiao/test9-1.cpp
new file mode 100644
index 0000000..668f2e2
--- /dev/null
+++ b/14lanqiao/test9-1.cpp
@@ -0,0 +1,60 @@
+// lanqiao3516 景区导游(floyd)
+#include<bits/stdc++.h>
+using namespace std;
+const int N = 8e3 + 10, INF = INT_MAX;
+
+bool vis[N]; // 标记数组
+int d[N][N]; // 最段路数组
+int n, m;
+// floyd 求多源最短路 - 时间复杂度-O(n^3)
+void floyd(){
+    // 枚举中转点
+    for(int k = 1; k <= n; k++){
+        // 枚举任意两点
+        for(int i = 1; i <= n; i++){
+            for(int j = 1; j <= n; j++){
+                if(d[i][k] != INF && d[k][j] != INF){
+                    if(d[i][k] + d[k][j] < d[i][j]){
+                        d[i][j] = d[i][k] + d[k][j];
+                    }
+                }
+            }
+        }
+    }
+}
+int main(){
+    fill(d[0], d[0] + N * N, INF);
+    cin >> n >> m;
+
+    for(int i = 1; i <= n - 1; i++){
+        int u, v, w; cin >> u >> v >> w;
+        d[u][v] = d[v][u] = w;
+    }
+    vector<int> p;
+    for(int i = 1; i <= m; i++){
+        int v; cin >> v;
+        p.push_back(v);
+    }
+    floyd();
+    vector<int> vec = p;
+    for(int i = 0; i < p.size(); i++){
+        vec.erase(vec.begin() + i);
+        int ans = 0;
+        for(int j = 0; j < vec.size() - 1; j++){
+            int v = vec[j], u = vec[j+1];
+            ans += d[u][v];
+        }
+        cout << ans << ' ';
+        vec = p;
+    }
+    return 0;
+}
+/* test sanples -> 10 7 13 14
+6 4
+1 2 1
+1 3 1
+3 4 2
+3 5 2
+4 6 3
+2 6 5 1
+*/
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