From ffbfc9e2731735ff48e15d6054995efc79b0108d Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BB=84=E6=B5=B7?= <10402852@qq.com>
Date: Wed, 13 Mar 2024 15:41:42 +0800
Subject: [PATCH] 'commit'

---
 TangDou/AcWing/BeiBao/1013.md | 20 +++++++++-----------
 1 file changed, 9 insertions(+), 11 deletions(-)

diff --git a/TangDou/AcWing/BeiBao/1013.md b/TangDou/AcWing/BeiBao/1013.md
index e847225..24632dc 100644
--- a/TangDou/AcWing/BeiBao/1013.md
+++ b/TangDou/AcWing/BeiBao/1013.md
@@ -343,13 +343,13 @@ int n;
 int m;
 int path[N], res[N];
 int w[N][M];
-int Max;
+int mx;
 
 // u:第几个公司　s:已经产生的价值 r:剩余的机器数量
 void dfs(int u, int s, int r) {
     if (u == n + 1) {
-        if (s > Max) {
-            Max = s;
+        if (s > mx) {
+            mx = s;
             memcpy(res, path, sizeof path);
         }
         return;
@@ -357,25 +357,23 @@ void dfs(int u, int s, int r) {
 
     for (int i = 0; i <= r; i++) {
         path[u] = i;
-        dfs(u + 1, s + w[u][i], r - i);//给u号公司分配i个机器
-        // path[u] = 0;
-        //按照回溯法，此处应该写path[u]还原现场，但本题中即使不还原现场，path[u]也会被下一次循环所覆盖，所以这句可以省略掉
+        dfs(u + 1, s + w[u][i], r - i); // 给u号公司分配i个机器
+        path[u] = 0;
     }
 }
 
 int main() {
-    scanf("%d %d", &n, &m);
+    cin >> n >> m;
 
     for (int i = 1; i <= n; i++)
         for (int j = 1; j <= m; j++)
-            scanf("%d", &w[i][j]);
+            cin >> w[i][j];
 
     dfs(1, 0, m);
 
-    printf("%d\n", Max);
-    //输出最优答案时的路径
+    printf("%d\n", mx);
+    // 输出最优答案时的路径
     for (int i = 1; i <= n; i++) printf("%d %d\n", i, res[i]);
     return 0;
 }
-
 ```
\ No newline at end of file