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‎README.md

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通知：开始更新图论内容，图论部分还没有其他语言版本，欢迎录友们提交PR，成为contributor
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### 深搜广搜
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* [图论：深度优先搜索理论基础](./problems/图论深搜理论基础.md)
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* [图论：797.所有可能的路径](./problems/0797.所有可能的路径.md)
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* [图论：广度优先搜索理论基础](./problems/图论广搜理论基础.md)
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* [图论：200.岛屿数量.深搜版](./problems/0200.岛屿数量.深搜版.md)
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* [图论：200.岛屿数量.广搜版](./problems/0200.岛屿数量.广搜版.md)
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* [图论：695.岛屿的最大面积](./problems/0695.岛屿的最大面积.md)
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* [图论：1020.飞地的数量](./problems/1020.飞地的数量.md)
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* [图论：130.被围绕的区域](./problems/0130.被围绕的区域.md)
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* [图论：417.太平洋大西洋水流问题](./problems/0417.太平洋大西洋水流问题.md)
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* [图论：827.最大人工岛](./problems/0827.最大人工岛.md)
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* [图论：127. 单词接龙](./problems/0127.单词接龙.md)
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* [图论：841.钥匙和房间](./problems/0841.钥匙和房间.md)
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* [图论：463. 岛屿的周长](./problems/0463.岛屿的周长.md)
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* [图论：并查集理论基础](./problems/图论并查集理论基础.md)
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* [图论：1971. 寻找图中是否存在路径](./problems/1971.寻找图中是否存在路径.md)
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* [图论：684.冗余连接](./problems/0684.冗余连接.md)
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* [图论：685.冗余连接II](./problems/0685.冗余连接II.md)
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1. [图论：理论基础](./problems/kamacoder/图论理论基础.md)
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2. [图论：深度优先搜索理论基础](./problems/kamacoder/图论深搜理论基础.md)
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3. [图论：所有可达路径](./problems/kamacoder/0098.所有可达路径.md)
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4. [图论：广度优先搜索理论基础](./problems/kamacoder/图论广搜理论基础.md)
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5. [图论：岛屿数量.深搜版](./problems/kamacoder/0099.岛屿的数量深搜.md)
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6. [图论：岛屿数量.广搜版](./problems/kamacoder/0099.岛屿的数量广搜.md)
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7. [图论：岛屿的最大面积](./problems/kamacoder/0100.岛屿的最大面积.md)
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8. [图论：孤岛的总面积](./problems/kamacoder/0101.孤岛的总面积.md)
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9. [图论：沉没孤岛](./problems/kamacoder/0102.沉没孤岛.md)
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10. [图论：水流问题](./problems/kamacoder/0103.水流问题.md)
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11. [图论：建造最大岛屿](./problems/kamacoder/0104.建造最大岛屿.md)
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12. [图论：字符串接龙](./problems/kamacoder/0110.字符串接龙.md)
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13. [图论：有向图的完全可达性](./problems/kamacoder/0105.有向图的完全可达性.md)
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14. [图论：岛屿的周长](./problems/kamacoder/0106.岛屿的周长.md)
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15. [图论：并查集理论基础](./problems/kamacoder/图论并查集理论基础.md)
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16. [图论：寻找存在的路径](./problems/kamacoder/0107.寻找存在的路径.md)
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17. [图论：冗余连接](./problems/kamacoder/0108.冗余连接.md)
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18. [图论：冗余连接II](./problems/kamacoder/0109.冗余连接II.md)
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19. [图论：最小生成树之prim](./problems/kamacoder/0053.寻宝-prim.md)
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20. [图论：最小生成树之kruskal](./problems/kamacoder/0053.寻宝-Kruskal.md)
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21. [图论：拓扑排序](./problems/kamacoder/0117.软件构建.md)
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22. [图论：dijkstra（朴素版）](./problems/kamacoder/0047.参会dijkstra朴素.md)
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23. [图论：dijkstra（堆优化版）](./problems/kamacoder/0047.参会dijkstra堆.md)
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24. [图论：Bellman_ford 算法](./problems/kamacoder/0094.城市间货物运输I.md)
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25. [图论：Bellman_ford 队列优化算法（又名SPFA）](./problems/kamacoder/0094.城市间货物运输I-SPFA.md)
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26. [图论：Bellman_ford之判断负权回路](./problems/kamacoder/0095.城市间货物运输II.md)
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27. [图论：Bellman_ford之单源有限最短路](./problems/kamacoder/0095.城市间货物运输II.md)
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28. [图论：Floyd 算法](./problems/kamacoder/0097.小明逛公园.md)
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29. [图论：A * 算法](./problems/kamacoder/0126.骑士的攻击astar.md)
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30. [图论：最短路算法总结篇](./problems/kamacoder/最短路问题总结篇.md)
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31. [图论：图论总结篇](./problems/kamacoder/图论总结篇.md)
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（持续更新中....）
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‎problems/0417.太平洋大西洋水流问题.md

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// 记录从大西洋出发，可以遍历的节点
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vector<vector<bool>> atlantic = vector<vector<bool>>(n, vector<bool>(m, false));
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// 从最上最下行的节点出发，向高处遍历
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// 从最左最右列的节点出发，向高处遍历
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for (int i = 0; i < n; i++) {
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dfs (heights, pacific, i, 0); // 遍历最左列，接触太平洋
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dfs (heights, atlantic, i, m - 1); // 遍历最右列，接触大西
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}
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// 从最左最右列的节点出发，向高处遍历
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// 从最上最下行的节点出发，向高处遍历
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for (int j = 0; j < m; j++) {
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dfs (heights, pacific, 0, j); // 遍历最上行，接触太平洋
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dfs (heights, atlantic, n - 1, j); // 遍历最下行，接触大西洋
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}
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```
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```Java
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class Solution {
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// 和Carl题解更加符合的Java DFS
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private int[][] directions = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
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/**
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* @param heights 题目给定的二维数组
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* @param m 当前位置的行号
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* @param n 当前位置的列号
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* @param visited 记录这个位置可以到哪条河
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*/
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public void dfs(int[][] heights, boolean[][] visited, int m, int n){
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if(visited[m][n]) return;
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visited[m][n] = true;
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for(int[] dir: directions){
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int nextm = m + dir[0];
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int nextn = n + dir[1];
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//出了2D array的边界，continue
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if(nextm < 0||nextm == heights.length||nextn <0||nextn== heights[0].length) continue;
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//下一个位置比当下位置还要低，跳过，继续找下一个更高的位置
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if(heights[m][n] > heights[nextm][nextn]) continue;
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dfs(heights, visited, nextm, nextn);
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}
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}
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public List<List<Integer>> pacificAtlantic(int[][] heights) {
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int m = heights.length;
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int n = heights[0].length;
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// 记录从太平洋边出发，可以遍历的节点
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boolean[][] pacific = new boolean[m][n];
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// 记录从大西洋出发，可以遍历的节点
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boolean[][] atlantic = new boolean[m][n];
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// 从最左最右列的节点出发，向高处遍历
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for(int i = 0; i<m; i++){
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dfs(heights, pacific, i, 0); //遍历pacific最左边
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dfs(heights, atlantic, i, n-1); //遍历atlantic最右边
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}
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// 从最上最下行的节点出发，向高处遍历
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for(int j = 0; j<n; j++){
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dfs(heights, pacific, 0, j); //遍历pacific最上边
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dfs(heights, atlantic, m-1, j); //遍历atlantic最下边
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}
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List<List<Integer>> result = new ArrayList<>();
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for(int a = 0; a<m; a++){
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for(int b = 0; b<n; b++){
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// 如果这个节点，从太平洋和大西洋出发都遍历过，就是结果
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if(pacific[a][b] && atlantic[a][b]){
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List<Integer> pair = new ArrayList<>();
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pair.add(a);
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pair.add(b);
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result.add(pair);
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}
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}
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}
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return result;
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}
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}
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```
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广度优先遍历：
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```Java

‎problems/0474.一和零.md

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* 时间复杂度: O(kmn)，k 为strs的长度
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* 空间复杂度: O(mn)
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C++:
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使用三维数组的版本
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```CPP
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class Solution {
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public:
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int findMaxForm(vector<string>& strs, int m, int n) {
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int num_of_str = strs.size();
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vector<vector<vector<int>>> dp(num_of_str, vector<vector<int>>(m + 1,vector<int>(n + 1, 0)));
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/* dp[i][j][k] represents, if choosing items among strs[0] to strs[i] to form a subset,
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what is the maximum size of this subset such that there are no more than m 0's and n 1's in this subset.
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Each entry of dp[i][j][k] is initialized with 0
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transition formula:
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using x[i] to indicates the number of 0's in strs[i]
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using y[i] to indicates the number of 1's in strs[i]
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dp[i][j][k] = max(dp[i-1][j][k], dp[i-1][j - x[i]][k - y[i]] + 1)
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*/
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// num_of_zeros records the number of 0's for each str
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// num_of_ones records the number of 1's for each str
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// find the number of 0's and the number of 1's for each str in strs
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vector<int> num_of_zeros;
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vector<int> num_of_ones;
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for (auto& str : strs){
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int count_of_zero = 0;
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int count_of_one = 0;
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for (char &c : str){
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if(c == '0') count_of_zero ++;
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else count_of_one ++;
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}
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num_of_zeros.push_back(count_of_zero);
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num_of_ones.push_back(count_of_one);
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}
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// num_of_zeros[0] indicates the number of 0's for str[0]
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// num_of_ones[0] indiates the number of 1's for str[1]
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// initialize the 1st plane of dp[i][j][k], i.e., dp[0][j][k]
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// if num_of_zeros[0] > m or num_of_ones[0] > n, no need to further initialize dp[0][j][k],
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// because they have been intialized to 0 previously
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if(num_of_zeros[0] <= m && num_of_ones[0] <= n){
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// for j < num_of_zeros[0] or k < num_of_ones[0], dp[0][j][k] = 0
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for(int j = num_of_zeros[0]; j <= m; j++){
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for(int k = num_of_ones[0]; k <= n; k++){
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dp[0][j][k] = 1;
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}
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}
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}
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/* if j - num_of_zeros[i] >= 0 and k - num_of_ones[i] >= 0:
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dp[i][j][k] = max(dp[i-1][j][k], dp[i-1][j - num_of_zeros[i]][k - num_of_ones[i]] + 1)
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else:
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dp[i][j][k] = dp[i-1][j][k]
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*/
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for (int i = 1; i < num_of_str; i++){
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int count_of_zeros = num_of_zeros[i];
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int count_of_ones = num_of_ones[i];
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for (int j = 0; j <= m; j++){
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for (int k = 0; k <= n; k++){
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if( j < count_of_zeros || k < count_of_ones){
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dp[i][j][k] = dp[i-1][j][k];
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}else{
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dp[i][j][k] = max(dp[i-1][j][k], dp[i-1][j - count_of_zeros][k - count_of_ones] + 1);
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}
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}
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}
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}
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return dp[num_of_str-1][m][n];
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}
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};
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```
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## 总结
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‎problems/kamacoder/0047.参会dijkstra堆.md

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<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>
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# dijkstra（堆优化版）精讲
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[卡码网：47. 参加科学大会](https://kamacoder.com/problempage.php?pid=1047)

‎problems/kamacoder/0047.参会dijkstra朴素.md

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<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>
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# dijkstra（朴素版）精讲
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[卡码网：47. 参加科学大会](https://kamacoder.com/problempage.php?pid=1047)

‎problems/kamacoder/0053.寻宝-Kruskal.md

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<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>
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# kruskal算法精讲
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[卡码网：53. 寻宝](https://kamacoder.com/problempage.php?pid=1053)

‎problems/kamacoder/0053.寻宝-prim.md

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<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>
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# prim算法精讲
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[卡码网：53. 寻宝](https://kamacoder.com/problempage.php?pid=1053)

‎problems/kamacoder/0054.替换数字.md

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<p align="center">
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<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">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收益！</strong></p>
7-
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# 替换数字
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[卡码网题目链接](https://kamacoder.com/problempage.php?pid=1064)
@@ -376,9 +369,3 @@ main();
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### Rust:
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<p align="center">
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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‎problems/kamacoder/0055.右旋字符串.md

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<p align="center">
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<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">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收益！</strong></p>
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# 右旋字符串
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### Rust:
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<p align="center">
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>

‎problems/kamacoder/0094.城市间货物运输I-SPFA.md

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<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>
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# Bellman_ford 队列优化算法（又名SPFA）
35

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[卡码网：94. 城市间货物运输 I](https://kamacoder.com/problempage.php?pid=1152)

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