From 96c6bdaa522ad073d1890dbe28ccf2243c9edd40 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BB=84=E6=B5=B7?= <10402852@qq.com>
Date: Wed, 3 Jan 2024 14:27:52 +0800
Subject: [PATCH] 'commit'

---
 TangDou/AcWing_TiGao/T3/Floyd/344.cpp | 14 +++++++-------
 1 file changed, 7 insertions(+), 7 deletions(-)

diff --git a/TangDou/AcWing_TiGao/T3/Floyd/344.cpp b/TangDou/AcWing_TiGao/T3/Floyd/344.cpp
index 78ccb7c..e4cee81 100644
--- a/TangDou/AcWing_TiGao/T3/Floyd/344.cpp
+++ b/TangDou/AcWing_TiGao/T3/Floyd/344.cpp
@@ -4,7 +4,7 @@ using namespace std;
 const int N = 110, INF = 0x3f3f3f3f;
 int n, m;
 int g[N][N], dis[N][N];
-int path[N], idx;
+vector<int> path;
 int mid[N][N];
 int ans = INF;
 
@@ -13,7 +13,7 @@ void get_path(int i, int j) {
     int k = mid[i][j]; // 获取中间转移点
     if (!k) return;    // 如果i,j之间没有中间点，停止
     get_path(i, k);    // 递归前半段
-    path[idx++] = k;   // 记录k节点
+    path.push_back(k); // 记录k节点
     get_path(k, j);    // 递归后半段
 }
 
@@ -54,11 +54,11 @@ int main() {
                     //  2. i->j 中间的路线不明，需要用get_path进行查询出i->j的最短路径怎么走,当然，也是在<k的范围内的
                     //  3. 记录j
                     //  4. 记录k
-                    idx = 0;
-                    path[idx++] = i;
+                    path.clear();
+                    path.push_back(i);
                     get_path(i, j); // i是怎么到达j的？就是问dis[i][j]是怎么获取到的，这是在求最短路径过程中的一个路径记录问题
-                    path[idx++] = j;
-                    path[idx++] = k;
+                    path.push_back(j);
+                    path.push_back(k);
                 }
 
         // 正常floyd
@@ -73,7 +73,7 @@ int main() {
     if (ans == INF)
         puts("No solution.");
     else
-        for (int i = 0; i < idx; i++) cout << path[i] << ' ';
+        for (int i = 0; i < path.size(); i++) cout << path[i] << ' ';
 
     return 0;
 }
\ No newline at end of file