【LeetCode】拓扑排序

【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 };

View Code

【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 };

View Code

【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 };

View Code

【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 };