update font color

data/1.算法初阶/7.蓝桥杯-动态规划/6.糖果/config.json

+{
+  "node_id": "algorithm-30a892869c304cd483969480fe88b106",
+  "keywords": [
+    "蓝桥杯",
+    "糖果"
+  ],
+  "children": [],
+  "export": []
+}
\ No newline at end of file

data/1.算法初阶/7.蓝桥杯-动态规划/6.糖果/desc.md

+糖果店的老板一共有M种口味的糖果出售。为了方便描述，我们将M种口味编号1~M。  
+小明希望能品尝到所有口味的糖果。遗憾的是老板并不单独出售糖果，而是K颗一包整包出售。  
+幸好糖果包装上注明了其中K颗糖果的口味，所以小明可以在买之前就知道每包内的糖果口味。  
+给定包糖果，请你计算小明最少买几包，就可以品尝到所有口味的糖果。  
+#### 输入
+第一行包含三个整数N、M 和K。  
+接下来N 行每行K 这整数$T_1,T_2,…,T_K$，代表一包糖果的口味。  
+1<=N<=100，1<=M<=20，1<=K<=20，1<=$T_i$<=M。  
+#### 输出
+一个整数表示答案。如果小明无法品尝所有口味，输出-1。
+
+#### 样例输入
+```
+6 5 3
+1 1 2
+1 2 3
+1 1 3
+2 3 5
+5 4 2
+5 1 2
+```
+#### 样例输出
+```
+2
+```
\ No newline at end of file

data/1.算法初阶/7.蓝桥杯-动态规划/6.糖果/solution.cpp

+#include <bits/stdc++.h>
+using namespace std;
+int dp[1 << 20]; //dp[v]表示口味为v时所需要的最少糖果包数
+int ST[100];     //100包糖果
+int main()
+{
+    int n, m, k;
+    cin >> n >> m >> k;         //输入n包糖果，m个口味，每包k个口味
+    int tot = (1 << m) - 1;     //表示所有m种口味各种组合方式
+    memset(dp, -1, sizeof dp);  //初始化dp全部为-1
+    for (int i = 0; i < n; i++) //依次处理n包糖果
+    {
+        int st = 0;
+        for (int j = 0; j < k; j++) //依次处理每颗糖果口味
+        {
+            int x;
+            cin >> x;
+            st |= (1 << x - 1); //把第i包的第j颗糖果加入该包的口味st中
+        }
+        dp[st] = 1;
+        ST[i] = st;
+    }
+    for (int i = 0; i <= tot; i++) //遍历所有口味组合方式
+    {
+        if (dp[i] != -1) //表示已存在得到该口味的最少糖果包数量
+        {
+            for (int j = 0; j < n; j++) //检查给定的n包糖果
+            {
+                int st = ST[j];
+                if (dp[i | st] == -1 || dp[i | st] > dp[i] + 1) //状态转移
+                    dp[i | st] = dp[i] + 1;
+            }
+        }
+    }
+    cout << dp[tot]; //得到所有口味tot的最少糖果包数
+    return 0;
+}

data/1.算法初阶/7.蓝桥杯-动态规划/6.糖果/solution.json

+{
+  "type": "code_options",
+  "author": "CSDN.net",
+  "source": "solution.md",
+  "exercise_id": "3c7880dd590c4f7cbe182c2c00af0e5a"
+}
\ No newline at end of file

data/1.算法初阶/7.蓝桥杯-动态规划/6.糖果/solution.md

+# 糖果
+
+糖果店的老板一共有M种口味的糖果出售。为了方便描述，我们将M种口味编号1~M。  
+小明希望能品尝到所有口味的糖果。遗憾的是老板并不单独出售糖果，而是K颗一包整包出售。  
+幸好糖果包装上注明了其中K颗糖果的口味，所以小明可以在买之前就知道每包内的糖果口味。  
+给定包糖果，请你计算小明最少买几包，就可以品尝到所有口味的糖果。  
+
+**输入**
+
+第一行包含三个整数N、M 和K。  
+接下来N 行每行K 这整数$T_1,T_2,…,T_K$，代表一包糖果的口味。  
+1<=N<=100，1<=M<=20，1<=K<=20，1<=$T_i$<=M。  
+
+**输出**
+
+一个整数表示答案。如果小明无法品尝所有口味，输出-1。
+
+
+**样例输入**
+
+```
+6 5 3
+1 1 2
+1 2 3
+1 1 3
+2 3 5
+5 4 2
+5 1 2
+```
+
+**样例输出**
+
+```
+2
+```
+
+以下错误的一项是？
+
+## aop
+
+### before
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+```
+
+### after
+
+```cpp
+
+```
+
+## 答案
+
+```cpp
+const int N = 105, M = (1 << 20) + 10;
+int n, m, k, x, a[N], dp[M];
+
+int main()
+{
+    ios::sync_with_stdio(false);
+    cin >> n >> m >> k;
+    memset(dp, -1, sizeof(dp));
+    for (int i = 1; i <= n; i++)
+    {
+        int s = 0;
+        for (int j = 1; j <= k; j++)
+        {
+            cin >> x;
+            s |= (1 << (x - 1));
+        }
+        dp[s] = 1;
+        a[i] = s;
+    }
+    for (int i = 1; i <= n; i++)
+    {
+        for (int j = 0; j < (1 << m); j++)
+        {
+            if (dp[j] == -1)
+                continue;
+            int to = j | a[i];
+            if (dp[to] != -1)
+            {
+                dp[to] = min(dp[to], dp[j]);
+            }
+            else
+            {
+                dp[to] = dp[j] + 1;
+            }
+        }
+    }
+    printf("%d\n", dp[(1 << m) - 1]);
+    return 0;
+}
+```
+
+## 选项
+
+### A
+
+```cpp
+int n, m, k;
+int dp[1 << 20];
+vector<int> a;
+
+int main()
+{
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+    cin >> n >> m >> k;
+    for (int i = 0; i < (1 << m); i++)
+    {
+        dp[i] = 9999;
+    }
+    for (int i = 0; i < n; i++)
+    {
+        int s = 0;
+        for (int i = 0; i < k; i++)
+        {
+            int temp;
+            cin >> temp;
+            s |= 1 << (temp - 1);
+        }
+        a.push_back(s);
+    }
+    dp[0] = 0;
+    for (int i = 0; i < n; i++)
+    {
+        for (int j = 0; j < (1 << m); j++)
+        {
+            if (dp[j] == 9999 || (j | a[i]) == j)
+                continue;
+            dp[j | a[i]] = min(dp[j] + 1, dp[j | a[i]]);
+        }
+    }
+    if (dp[(1 << m) - 1] == 9999)
+        cout << -1;
+    else
+        cout << dp[(1 << m) - 1];
+    return 0;
+}
+```
+
+### B
+
+```cpp
+const int maxn = (1 << 20) + 52;
+const int inf = 9;
+int a[505], n, m, k, dp[2][maxn];
+int min(int a, int b, int c)
+{
+    return min(min(a, b), c);
+}
+int main()
+{
+    scanf("%d%d%d", &n, &m, &k);
+    int e = 1 << m;
+
+    for (int i = 1; i <= n; i++)
+    {
+        a[i] = 0;
+        for (int j = 0; j < k; j++)
+        {
+            int temp;
+            scanf("%d", &temp);
+            a[i] = a[i] | 1 << (temp - 1);
+        }
+    }
+
+    for (int i = 0; i < maxn; i++)
+    {
+        dp[0][i] = dp[1][i] = inf;
+    }
+    dp[0][0] = 0;
+
+    bool pos = 1;
+    for (int i = 1; i <= n; i++, pos = !pos)
+    {
+        for (int j = 0; j < e; j++)
+        {
+            dp[pos][j] = dp[!pos][j];
+        }
+        for (int j = 0; j < e; j++)
+        {
+            dp[pos][j | a[i]] = min(dp[pos][j | a[i]], dp[!pos][j | a[i]], dp[!pos][j] + 1);
+        }
+    }
+
+    if (dp[!pos][e - 1] == inf)
+    {
+        printf("-1\n");
+    }
+    else
+        printf("%d\n", dp[!pos][e - 1]);
+    return 0;
+}
+```
+
+### C
+
+```cpp
+int dp[1 << 20];
+int ST[100];
+int main()
+{
+    int n, m, k;
+    cin >> n >> m >> k;
+    int tot = (1 << m) - 1;
+    memset(dp, -1, sizeof dp);
+    for (int i = 0; i < n; i++)
+    {
+        int st = 0;
+        for (int j = 0; j < k; j++)
+        {
+            int x;
+            cin >> x;
+            st |= (1 << x - 1);
+        }
+        dp[st] = 1;
+        ST[i] = st;
+    }
+    for (int i = 0; i <= tot; i++)
+    {
+        if (dp[i] != -1)
+        {
+            for (int j = 0; j < n; j++)
+            {
+                int st = ST[j];
+                if (dp[i | st] == -1 || dp[i | st] > dp[i] + 1)
+                    dp[i | st] = dp[i] + 1;
+            }
+        }
+    }
+    cout << dp[tot];
+    return 0;
+}
+```

data_backup/3.蓝桥杯/37.糖果/solution.md

@@ -34,12 +34,15 @@
 2
 ```
 
+以下错误的一项是？
+
 ## aop
 
 ### before
 
 ```cpp
-
+#include <bits/stdc++.h>
+using namespace std;
 ```
 
 ### after
@@ -51,7 +54,45 @@
 ## 答案
 
 ```cpp
-
+const int N = 105, M = (1 << 20) + 10;
+int n, m, k, x, a[N], dp[M];
+
+int main()
+{
+    ios::sync_with_stdio(false);
+    cin >> n >> m >> k;
+    memset(dp, -1, sizeof(dp));
+    for (int i = 1; i <= n; i++)
+    {
+        int s = 0;
+        for (int j = 1; j <= k; j++)
+        {
+            cin >> x;
+            s |= (1 << (x - 1));
+        }
+        dp[s] = 1;
+        a[i] = s;
+    }
+    for (int i = 1; i <= n; i++)
+    {
+        for (int j = 0; j < (1 << m); j++)
+        {
+            if (dp[j] == -1)
+                continue;
+            int to = j | a[i];
+            if (dp[to] != -1)
+            {
+                dp[to] = min(dp[to], dp[j]);
+            }
+            else
+            {
+                dp[to] = dp[j] + 1;
+            }
+        }
+    }
+    printf("%d\n", dp[(1 << m) - 1]);
+    return 0;
+}
 ```
 
 ## 选项
@@ -59,17 +100,139 @@
 ### A
 
 ```cpp
-
+int n, m, k;
+int dp[1 << 20];
+vector<int> a;
+
+int main()
+{
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+    cin >> n >> m >> k;
+    for (int i = 0; i < (1 << m); i++)
+    {
+        dp[i] = 9999;
+    }
+    for (int i = 0; i < n; i++)
+    {
+        int s = 0;
+        for (int i = 0; i < k; i++)
+        {
+            int temp;
+            cin >> temp;
+            s |= 1 << (temp - 1);
+        }
+        a.push_back(s);
+    }
+    dp[0] = 0;
+    for (int i = 0; i < n; i++)
+    {
+        for (int j = 0; j < (1 << m); j++)
+        {
+            if (dp[j] == 9999 || (j | a[i]) == j)
+                continue;
+            dp[j | a[i]] = min(dp[j] + 1, dp[j | a[i]]);
+        }
+    }
+    if (dp[(1 << m) - 1] == 9999)
+        cout << -1;
+    else
+        cout << dp[(1 << m) - 1];
+    return 0;
+}
 ```
 
 ### B
 
 ```cpp
-
+const int maxn = (1 << 20) + 52;
+const int inf = 9;
+int a[505], n, m, k, dp[2][maxn];
+int min(int a, int b, int c)
+{
+    return min(min(a, b), c);
+}
+int main()
+{
+    scanf("%d%d%d", &n, &m, &k);
+    int e = 1 << m;
+
+    for (int i = 1; i <= n; i++)
+    {
+        a[i] = 0;
+        for (int j = 0; j < k; j++)
+        {
+            int temp;
+            scanf("%d", &temp);
+            a[i] = a[i] | 1 << (temp - 1);
+        }
+    }
+
+    for (int i = 0; i < maxn; i++)
+    {
+        dp[0][i] = dp[1][i] = inf;
+    }
+    dp[0][0] = 0;
+
+    bool pos = 1;
+    for (int i = 1; i <= n; i++, pos = !pos)
+    {
+        for (int j = 0; j < e; j++)
+        {
+            dp[pos][j] = dp[!pos][j];
+        }
+        for (int j = 0; j < e; j++)
+        {
+            dp[pos][j | a[i]] = min(dp[pos][j | a[i]], dp[!pos][j | a[i]], dp[!pos][j] + 1);
+        }
+    }
+
+    if (dp[!pos][e - 1] == inf)
+    {
+        printf("-1\n");
+    }
+    else
+        printf("%d\n", dp[!pos][e - 1]);
+    return 0;
+}
 ```
 
 ### C
 
 ```cpp
-
+int dp[1 << 20];
+int ST[100];
+int main()
+{
+    int n, m, k;
+    cin >> n >> m >> k;
+    int tot = (1 << m) - 1;
+    memset(dp, -1, sizeof dp);
+    for (int i = 0; i < n; i++)
+    {
+        int st = 0;
+        for (int j = 0; j < k; j++)
+        {
+            int x;
+            cin >> x;
+            st |= (1 << x - 1);
+        }
+        dp[st] = 1;
+        ST[i] = st;
+    }
+    for (int i = 0; i <= tot; i++)
+    {
+        if (dp[i] != -1)
+        {
+            for (int j = 0; j < n; j++)
+            {
+                int st = ST[j];
+                if (dp[i | st] == -1 || dp[i | st] > dp[i] + 1)
+                    dp[i | st] = dp[i] + 1;
+            }
+        }
+    }
+    cout << dp[tot];
+    return 0;
+}
 ```

leetcode_helper.py

@@ -457,8 +457,8 @@ def add_color_for_special_exercises():
                 print(solution_md_path)
                 with open(solution_md_path, 'r', encoding='utf-8') as f:
                     solution_md_data = f.read()
-                if solution_md_data.find('<font color="red">错误</font>') == -1:
-                    solution_md_data = solution_md_data.replace('错误', '<font color="red">错误</font>')
+                if solution_md_data.find('<span style="color:red">错误</span>') == -1:
+                    solution_md_data = solution_md_data.replace('错误', '<span style="color:red">错误</span>')
                     with open(solution_md_path, 'w', encoding='utf-8') as f:
                         f.write(solution_md_data)