【207】 Course Schedule
排课问题，n门课排课，有的课程必须在另外一些课程之前上，问能不能排出来顺序。
题解：裸的拓扑排序。参考代码见算法竞赛入门指南这本书。
1 class Solution { 2 public: 3 bool dfs(const vector<vector<int>>& g, vector<int>& c, int u) { 4 c[u] = -1; 5 for (int v = 0; v < n; ++v) { 6 if (g[u][v]) { 7 if (c[v] < 0) { return false; } 8 else if (!c[v] && !dfs(g, c, v)) { 9 return false; 10 } 11 } 12 } 13 c[u] = 1; 14 topo[--t] = u; 15 return true; 16 } 17 bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) { 18 vector<vector<int>> graph(numCourses, vector<int>(numCourses, 0)); 19 n = numCourses; 20 topo.resize(n); 21 t = n; 22 for (auto ele : prerequisites) { 23 int u = ele.first, v = ele.second; 24 graph[v][u] = 1; 25 } 26 vector<int> c(n, 0); 27 for (int i = 0; i < n; ++i) { 28 if (!c[i]) { 29 if (!dfs(graph, c, i)) { 30 return false; 31 } 32 } 33 } 34 /* 35 for (int i = 0; i < n; ++i) { 36 cout << topo[i] << " " ; 37 } 38 cout << endl; 39 */ 40 return true; 41 } 42 vector<int> topo; 43 int n, t; 44 };
【210】 Course Schedule II
同上一个排课问题，这次的问题是能不能给出一个可行的顺序。
题解：还是裸的拓扑排序。
1 class Solution { 2 public: 3 bool dfs(vector<int>& c, vector<int>& topo, int u) { 4 c[u] = -1; 5 for (int v = 0; v < n; ++v) { 6 if (g[u][v]) { 7 if (c[v] < 0) {return false;} 8 else if (!c[v] && !dfs(c, topo, v)) {return false; } 9 } 10 } 11 c[u] = 1; 12 topo[--t] = u; 13 return true; 14 } 15 vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) { 16 n = numCourses, t = n; 17 vector<int> topo(n, 0); 18 vector<int> c(n, 0); 19 vector<vector<int>> graph(n, vector<int>(n, 0)); 20 for (auto ele : prerequisites) { 21 int u = ele.first, v = ele.second; 22 graph[v][u] = 1; 23 } 24 g = graph; 25 26 for (int u = 0; u < n; ++u) { 27 if (!c[u]) { 28 if (!dfs(c, topo, u)) { 29 vector<int> temp; 30 return temp; 31 } 32 } 33 } 34 return topo; 35 } 36 int n, t; 37 vector<vector<int>> g; 38 };
【269】 Alien Dictionary
给了一门新的语言，给了一个单词字典，所有的单词按照字典序排序。要求返回现有字母的顺序，没有顺序的话，返回空数组。
题解：逐个比较两个相邻的单词，如果他们第i个位置不同，说明前一个单词的第i个字母u，要小于后一个单词的第i个字母v，然后建图，建完图直接裸的拓扑排序。
1 class Solution { 2 public: 3 bool dfs(int u) { 4 c[u] = -1; 5 for (int v = 0; v < tot; ++v) { 6 if (g[u][v]) { 7 if (c[v] < 0) {return false;} 8 else if (!c[v] && !dfs(v)) {return false;} 9 } 10 } 11 c[u] = 1; 12 topo[--cur] = u; 13 return true; 14 } 15 16 string alienOrder(vector<string>& words) { 17 vector<pair<int, int>> order; 18 const int n = words.size(); 19 int t = 0; 20 for (int i = 0; i < n; ++i) { 21 string word = words[i]; 22 for (auto ele : word) { 23 if (mpCh2Num.find(ele) == mpCh2Num.end()) { 24 mpCh2Num[ele] = t; 25 mpNum2Ch[t] = ele; 26 ++t; 27 } 28 } 29 } 30 c.resize(t), topo.resize(t); 31 tot = t; cur = t; 32 33 for (int i = 0; i < n - 1; ++i) { 34 string word1 = words[i], word2 = words[i+1]; 35 for (int idx = 0; idx < min(word1.size(), word2.size()); ++idx) { 36 if (word1[idx] != word2[idx]) { 37 pair<int, int> p = make_pair(mpCh2Num[word1[idx]], mpCh2Num[word2[idx]]); 38 order.push_back(p); 39 break; 40 } 41 } 42 } 43 44 vector<vector<int>> graph(t, vector<int>(t, 0)); 45 for (auto ele : order) { 46 int u = ele.first, v = ele.second; 47 graph[u][v] = 1; 48 } 49 g = graph; 50 51 for (int u = 0; u < t; ++u) { 52 if (!c[u]) { 53 if (!dfs(u)) { 54 string temp; 55 return temp; 56 } 57 } 58 } 59 string ans; 60 for (auto ele : topo) { 61 ans += mpNum2Ch[ele]; 62 } 63 return ans; 64 } 65 vector<vector<int>> g; 66 vector<int> c, topo; 67 map<int, char> mpNum2Ch; 68 map<char, int> mpCh2Num; 69 int tot; 70 int cur; 71 };
【329】 Longest Increasing Path in a Matrix
给了一个矩阵matrix， 一个点他可以朝着上下左右四个方向走，问这个矩阵能走出来的最长递增的路径的长度是多少。
题解：裸的dfs会超时，所以加上了一个记忆化数组过了。题目的解法三有拓扑排序的相关解法，下次要搞懂那个解法。
1 class Solution { 2 public: 3 void print(vector<vector<int>>& mat) { 4 const int n = mat.size(), m = mat[0].size(); 5 for (int i = 0; i < n; ++i) { 6 for (int j = 0; j < m; ++j) { 7 cout << mat[i][j] << " "; 8 } 9 cout << endl; 10 } 11 } 12 int dirx[4] = {-1, 0, 1, 0}; 13 int diry[4] = {0, -1, 0, 1}; 14 int dfs(const vector<vector<int>>& mat, int x, int y, vector<vector<int>>& vis) { 15 vis[x][y] = 1; 16 for (int i = 0; i < 4; ++i) { 17 int newx = x + dirx[i], newy = y + diry[i]; 18 if (newx >= 0 && newx < n && newy >= 0 && newy < m && !vis[newx][newy]&& mat[newx][newy] > mat[x][y]) { 19 if (memo[newx][newy] != 0) { 20 memo[x][y] = max(memo[x][y], memo[newx][newy] + 1); 21 } else { 22 memo[x][y] = max(memo[x][y], dfs(mat, newx, newy, vis) + 1); 23 } 24 } 25 } 26 vis[x][y] = 0; 27 return memo[x][y]; 28 } 29 int longestIncreasingPath(vector<vector<int>>& matrix) { 30 n = matrix.size(); 31 if (n == 0) { return 0; } 32 m = matrix[0].size(); 33 if (m == 0) { return 0; } 34 35 int ans = 0; 36 memo = matrix; 37 for (int i = 0; i < n; ++i) { 38 for (int j = 0; j < m; ++j) { 39 memo[i][j] = 0; 40 } 41 } 42 43 for (int i = 0; i < n; ++i) { 44 for (int j = 0; j < m; ++j) { 45 vector<vector<int>> vis(n, vector<int>(m, 0)); 46 memo[i][j] = dfs(matrix, i, j, vis); 47 ans = max(ans, memo[i][j]); 48 } 49 } 50 return ans +1; 51 } 52 int n, m; 53 vector<vector<int>> memo; 54 };
【444】 Sequence Reconstruction