Update

youngyangyang04 · youngyangyang04 · commit 7aa973b56194 · 2025-07-06T11:28:34.000+08:00
diff --git a/problems/0042.接雨水.md b/problems/0042.接雨水.md
@@ -14,8 +14,7 @@
 
 示例 1：
 
-![](https://code-thinking-1253855093.cos.ap-guangzhou.myqcloud.com/pics/20210713205038.png)
-
+![image](https://file1.kamacoder.com/i/algo/2025-07-04_12-06-59.jpg) 
 * 输入：height = [0,1,0,2,1,0,1,3,2,1,2,1]
 * 输出：6
 * 解释：上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图，在这种情况下，可以接 6 个单位的雨水（蓝色部分表示雨水）。
diff --git a/problems/kamacoder/0044.开发商购买土地.md b/problems/kamacoder/0044.开发商购买土地.md
@@ -613,4 +613,3 @@ func main() {
 }
 ```
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0047.参会dijkstra堆.md b/problems/kamacoder/0047.参会dijkstra堆.md
@@ -927,4 +927,3 @@ func main() {
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0047.参会dijkstra朴素.md b/problems/kamacoder/0047.参会dijkstra朴素.md
@@ -942,4 +942,3 @@ main()
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0053.寻宝-Kruskal.md b/problems/kamacoder/0053.寻宝-Kruskal.md
@@ -760,4 +760,3 @@ int main()
 }
 
 ```
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0053.寻宝-prim.md b/problems/kamacoder/0053.寻宝-prim.md
@@ -757,4 +757,3 @@ main()
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0054.替换数字.md b/problems/kamacoder/0054.替换数字.md
@@ -432,4 +432,3 @@ echo $s;
 
 ### Rust:
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0055.右旋字符串.md b/problems/kamacoder/0055.右旋字符串.md
@@ -409,4 +409,3 @@ echo $s;
 
 ### Rust:
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0058.区间和.md b/problems/kamacoder/0058.区间和.md
@@ -408,4 +408,3 @@ func main() {
 }
 ```
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0094.城市间货物运输I-SPFA.md b/problems/kamacoder/0094.城市间货物运输I-SPFA.md
@@ -536,4 +536,3 @@ main()
 
 
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0094.城市间货物运输I.md b/problems/kamacoder/0094.城市间货物运输I.md
@@ -538,4 +538,3 @@ main()
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0095.城市间货物运输II.md b/problems/kamacoder/0095.城市间货物运输II.md
@@ -183,7 +183,6 @@ struct Edge { //邻接表
     Edge(int t, int w): to(t), val(w) {}  // 构造函数
 };
 
-
 int main() {
     int n, m, p1, p2, val;
     cin >> n >> m;
@@ -229,16 +228,111 @@ int main() {
             }
         }
     }
-
     if (flag) cout << "circle" << endl;
     else if (minDist[end] == INT_MAX) {
         cout << "unconnected" << endl;
     } else {
         cout << minDist[end] << endl;
     }
-
 }
+```
+
+以上代码看上去没问题，但提交之后会发现报错了，为什么呢？ 
+
+例如这组数据： 
+
+```
+4 6
+1 4 3
+1 2 1
+1 3 1
+3 2 -2
+2 4 1
+3 4 0
+```
+
+如图：
+
+![](https://file1.kamacoder.com/i/algo/2025-06-03_16-12-41.jpg)
+
+上面代码在执行的过程中，节点4 已经出队列了，但 计算入度会反复计算 节点4的入度，导致 误判 该图有环。
+
+我们需要记录哪些节点已经出队列了，哪些节点在队列里面，对于已经出队列的节点不用统计入度，以下为优化后的代码：
+
+```CPP 
+#include <iostream>
+#include <vector>
+#include <queue>
+#include <list>
+#include <climits>
+using namespace std;
+
+struct Edge { //邻接表
+    int to;  // 链接的节点
+    int val; // 边的权重
+
+    Edge(int t, int w): to(t), val(w) {}  // 构造函数
+};
+
+
+int main() {
+    int n, m, p1, p2, val;
+    cin >> n >> m;
+
+    vector<list<Edge>> grid(n + 1); // 邻接表
+
+    // 将所有边保存起来
+    for(int i = 0; i < m; i++){
+        cin >> p1 >> p2 >> val;
+        // p1 指向 p2，权值为 val
+        grid[p1].push_back(Edge(p2, val));
+    }
+    int start = 1;  // 起点
+    int end = n;    // 终点
+
+    vector<int> minDist(n + 1 , INT_MAX);
+    minDist[start] = 0;
+
+    queue<int> que;
+    que.push(start); // 队列里放入起点
+
+    vector<int> count(n+1, 0); // 记录节点加入队列几次
+    count[start]++;
+    vector<bool> inQueue(n+1, false);  // 记录节点是否在队列中
+
+    bool flag = false;
+    while (!que.empty()) {
+
+        int node = que.front(); que.pop();
+        inQueue[node] = false;  // 节点出队
+
+        for (Edge edge : grid[node]) {
+            int from = node;
+            int to = edge.to;
+            int value = edge.val;
+            if (minDist[to] > minDist[from] + value) { // 开始松弛
+                minDist[to] = minDist[from] + value;
+                 if (!inQueue[to]) {  // 避免重复入队
+                    que.push(to);
+                    inQueue[to] = true;
+                    count[to]++;
+                    if (count[to] == n) {// 如果加入队列次数超过 n-1次 就说明该图与负权回路
+                        flag = true;
+                        while (!que.empty()) que.pop();
+                        break;
+                    }
+                 }
+            }
+        }
+    }
 
+    if (flag) cout << "circle" << endl;
+    else if (minDist[end] == INT_MAX) {
+        cout << "unconnected" << endl;
+    } else {
+        cout << minDist[end] << endl;
+    }
+}
 ```
 
 ## 其他语言版本
@@ -456,4 +550,3 @@ if __name__ == "__main__":
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0096.城市间货物运输III.md b/problems/kamacoder/0096.城市间货物运输III.md
@@ -922,4 +922,3 @@ if __name__ == "__main__":
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0097.小明逛公园.md b/problems/kamacoder/0097.小明逛公园.md
@@ -574,4 +574,3 @@ if __name__ == '__main__':
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0098.所有可达路径.md b/problems/kamacoder/0098.所有可达路径.md
@@ -72,6 +72,7 @@
 * 1 <= N <= 100
 * 1 <= M <= 500
 
+**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[所有可达路径](https://www.bilibili.com/video/BV1VePeepEpP),相信结合视频再看本篇题解，更有助于大家对本题的理解**。
 
 ## 插曲  
 
@@ -887,4 +888,3 @@ async function dfs(graph, x, n) {
 
 
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0099.岛屿的数量广搜.md b/problems/kamacoder/0099.岛屿的数量广搜.md
@@ -46,6 +46,8 @@
 
 ## 思路 
 
+**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[图论：广搜有陷阱啊！！ | 广度优先搜索 | 卡码网：99.岛屿数量](https://www.bilibili.com/video/BV18PRGYcEiB/),相信结合视频再看本篇题解，更有助于大家对本题的理解**。
+
 注意题目中每座岛屿只能由**水平方向和/或竖直方向上**相邻的陆地连接形成。
 
 也就是说斜角度链接是不算了， 例如示例二，是三个岛屿，如图：
@@ -557,4 +559,3 @@ object Solution {
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0099.岛屿的数量深搜.md b/problems/kamacoder/0099.岛屿的数量深搜.md
@@ -46,6 +46,8 @@
 
 ## 思路 
 
+**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[图论：来用深搜解决一道题目，两种深搜写法，你掉坑了吗？ | 卡码网：99.岛屿数量](https://www.bilibili.com/video/BV18PRGYcEiB/),相信结合视频再看本篇题解，更有助于大家对本题的理解**。
+
 注意题目中每座岛屿只能由**水平方向和/或竖直方向上**相邻的陆地连接形成。
 
 也就是说斜角度链接是不算了， 例如示例二，是三个岛屿，如图：
@@ -459,4 +461,3 @@ object Solution {
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0100.岛屿的最大面积.md b/problems/kamacoder/0100.岛屿的最大面积.md
@@ -44,6 +44,8 @@
 
 ## 思路 
 
+**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[图论：深搜也有陷阱啊！！ | 深搜优先搜索 | 卡码网：100.岛屿的最大面积](https://www.bilibili.com/video/BV1FzoyY5EXH),相信结合视频再看本篇题解，更有助于大家对本题的理解**。
+
 注意题目中每座岛屿只能由**水平方向和/或竖直方向上**相邻的陆地连接形成。
 
 也就是说斜角度链接是不算了， 例如示例二，是三个岛屿，如图：
@@ -890,4 +892,3 @@ main();
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0101.孤岛的总面积.md b/problems/kamacoder/0101.孤岛的总面积.md
@@ -48,6 +48,8 @@
 
 ## 思路 
 
+**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[图论：岛屿问题再出新花样 | 深搜优先搜索 | 卡码网：101.孤岛总面积](https://www.bilibili.com/video/BV1mmZJYRESc),相信结合视频再看本篇题解，更有助于大家对本题的理解**。
+
 本题使用dfs，bfs，并查集都是可以的。 
 
 本题要求找到不靠边的陆地面积，那么我们只要从周边找到陆地然后 通过 dfs或者bfs 将周边靠陆地且相邻的陆地都变成海洋，然后再去重新遍历地图 统计此时还剩下的陆地就可以了。
@@ -700,4 +702,3 @@ const bfs = (graph, x, y) => {
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0102.沉没孤岛.md b/problems/kamacoder/0102.沉没孤岛.md
@@ -55,6 +55,8 @@
 
 ## 思路
 
+**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[图论：岛屿问题也需要逆向思考 | 深搜优先搜索DFS | 卡码网：102.沉没孤岛](https://www.bilibili.com/video/BV1fjdWYyEGu),相信结合视频再看本篇题解，更有助于大家对本题的理解**。
+
 这道题目和[0101.孤岛的总面积](https://kamacoder.com/problempage.php?pid=1173)正好反过来了，[0101.孤岛的总面积](https://kamacoder.com/problempage.php?pid=1173)是求 地图中间的空格数，而本题是要把地图中间的 1 都改成 0 。 
 
 那么两题在思路上也是差不多的。
@@ -503,4 +505,3 @@ const bfs = (graph, x, y) => {
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0103.水流问题.md b/problems/kamacoder/0103.水流问题.md
@@ -59,6 +59,8 @@
 
 ## 思路
 
+**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[图论：水流向何方？ | 深搜优先搜索DFS| 广度优先搜索BFS | 卡码网：103.水流问题](https://www.bilibili.com/video/BV1WNoEYrEio),相信结合视频再看本篇题解，更有助于大家对本题的理解**。
+
 一个比较直白的想法，其实就是 遍历每个点，然后看这个点 能不能同时到达第一组边界和第二组边界。
 
 至于遍历方式，可以用dfs，也可以用bfs，以下用dfs来举例。 
@@ -854,4 +856,3 @@ const isResult = (x, y) => {
 
 
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0104.建造最大岛屿.md b/problems/kamacoder/0104.建造最大岛屿.md
@@ -50,6 +50,8 @@
 
 ## 思路 
 
+**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[图论：岛屿问题上难度了！ |深搜优先搜索DFS | 广度优先搜索BFS | 卡码网：104.建造最大岛屿](https://www.bilibili.com/video/BV1Dn5CzZEw1/),相信结合视频再看本篇题解，更有助于大家对本题的理解**。
+
 本题的一个暴力想法，应该是遍历地图尝试 将每一个 0 改成1，然后去搜索地图中的最大的岛屿面积。 
 
 计算地图的最大面积：遍历地图 + 深搜岛屿，时间复杂度为 n * n。 
@@ -663,4 +665,3 @@ const dfs = (graph, visited, x, y, mark) => {
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0105.有向图的完全可达性.md b/problems/kamacoder/0105.有向图的完全可达性.md
@@ -1,7 +1,7 @@
 
 <p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
 
-# 105.有向图的完全可达性 
+# 105.有向图的完全联通
 
 [卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1177)
 
@@ -44,6 +44,8 @@
 
 ## 思路
 
+**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[图论：这次我们能到达有向图的所有节点吗？| 深度优先搜索 | 卡码网：105. 有向图的完全联通](https://www.bilibili.com/video/BV1QRJ6ziEvJ),相信结合视频再看本篇题解，更有助于大家对本题的理解**。
+
 本题给我们是一个有向图， 意识到这是有向图很重要！ 
 
 接下来我们再画一个图，从图里可以直观看出来，节点6 是 不能到达节点1 的
@@ -553,4 +555,3 @@ rl.on('close',()=>{
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0106.岛屿的周长.md b/problems/kamacoder/0106.岛屿的周长.md
@@ -47,6 +47,8 @@
 
 ## 思路
 
+**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[图论：这次的岛屿问题有点简单了。。 卡码网：106.岛屿的周长](https://www.bilibili.com/video/BV13fjczSEyV),相信结合视频再看本篇题解，更有助于大家对本题的理解**。
+
 岛屿问题最容易让人想到BFS或者DFS，但本题确实还用不上。 
 
 为了避免大家惯性思维，所以给大家安排了这道题目。 
@@ -359,4 +361,3 @@ func parseLine(line string, count int) []int {
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0107.寻找存在的路径.md b/problems/kamacoder/0107.寻找存在的路径.md
@@ -422,4 +422,3 @@ const isSame = (u, v) => {
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0108.冗余连接.md b/problems/kamacoder/0108.冗余连接.md
@@ -376,4 +376,3 @@ const isSame = (u, v) => {
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0109.冗余连接II.md b/problems/kamacoder/0109.冗余连接II.md
@@ -596,4 +596,3 @@ const getRemoveEdge = (edges) => {
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0110.字符串接龙.md b/problems/kamacoder/0110.字符串接龙.md
@@ -63,6 +63,8 @@ yhn
 
 ## 思路
 
+**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[图论：朴实无华的广搜这么难？ | 广度优先搜索 | 卡码网：110.字符串接龙](https://www.bilibili.com/video/BV1QEEizDEC4),相信结合视频再看本篇题解，更有助于大家对本题的理解**。
+
 以示例1为例，从这个图中可以看出 abc 到 def的路线 不止一条，但最短的一条路径上是4个节点。
 
 ![](https://file1.kamacoder.com/i/algo/20250317105155.png)
@@ -360,4 +362,3 @@ const init = async () => {
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0117.软件构建.md b/problems/kamacoder/0117.软件构建.md
@@ -539,4 +539,3 @@ const init = async () => {
 
 ### C
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/0126.骑士的攻击astar.md b/problems/kamacoder/0126.骑士的攻击astar.md
@@ -684,4 +684,3 @@ int main() {
 
 
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/图论为什么用ACM模式.md b/problems/kamacoder/图论为什么用ACM模式.md
@@ -91,4 +91,3 @@ cout << result[result.size() - 1];
 等大家将图论刷完，就会感受到我的良苦用心。加油
 
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/图论并查集理论基础.md b/problems/kamacoder/图论并查集理论基础.md
@@ -454,4 +454,3 @@ void join(int u, int v) {
 
 敬请期待 并查集题目精讲系列。
 
-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
diff --git a/problems/kamacoder/图论广搜理论基础.md b/problems/kamacoder/图论广搜理论基础.md
diff --git a/problems/kamacoder/图论总结篇.md b/problems/kamacoder/图论总结篇.md
diff --git a/problems/kamacoder/图论深搜理论基础.md b/problems/kamacoder/图论深搜理论基础.md
diff --git a/problems/kamacoder/图论理论基础.md b/problems/kamacoder/图论理论基础.md
diff --git a/problems/kamacoder/最短路问题总结篇.md b/problems/kamacoder/最短路问题总结篇.md
diff --git a/problems/qita/shejimoshi.md b/problems/qita/shejimoshi.md
diff --git a/problems/动态规划总结篇.md b/problems/动态规划总结篇.md

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`@@ -613,4 +613,3 @@ func main() {`
`613`	`613`	`}`
`614`	`614`	```
`615`	`615`
`616`		`-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>`
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`@@ -927,4 +927,3 @@ func main() {`
`927`	`927`
`928`	`928`	`### C`
`929`	`929`
`930`		`-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>`
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`@@ -942,4 +942,3 @@ main()`
`942`	`942`
`943`	`943`	`### C`
`944`	`944`
`945`		`-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>`
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`@@ -760,4 +760,3 @@ int main()`
`760`	`760`	`}`
`761`	`761`
`762`	`762`	```
`763`		`-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>`
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`@@ -757,4 +757,3 @@ main()`
`757`	`757`
`758`	`758`	`### C`
`759`	`759`
`760`		`-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>`
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`@@ -432,4 +432,3 @@ echo $s;`
`432`	`432`
`433`	`433`	`### Rust:`
`434`	`434`
`435`		`-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>`
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`@@ -409,4 +409,3 @@ echo $s;`
`409`	`409`
`410`	`410`	`### Rust:`
`411`	`411`
`412`		`-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>`
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`@@ -408,4 +408,3 @@ func main() {`
`408`	`408`	`}`
`409`	`409`	```
`410`	`410`
`411`		`-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>`
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`@@ -536,4 +536,3 @@ main()`
`536`	`536`
`537`	`537`
`538`	`538`
`539`		`-<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>`