leetcode Course Schedule
2015-12-08 09:30
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原题链接:https://leetcode.com/problems/course-schedule/
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?
For example:
2, [[1,0]]
There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.
2, [[1,0],[0,1]]
There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.
Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
拓扑排序。。
Description
There are a total of n courses you have to take, labeled from 0 to n−1.Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?
For example:
2, [[1,0]]
There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.
2, [[1,0],[0,1]]
There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.
Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
拓扑排序。。
class Solution { public: bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) { if (numCourses && prerequisites.empty()) return true; tot = 0; int num = 0, m = prerequisites.size(); inq = new int[numCourses + 10]; memset(inq, 0, sizeof(int)* (numCourses + 10)); head = new int[numCourses + 10]; memset(head, -1, sizeof(int)* (numCourses + 10)); G = new edge[m + 10]; for (int i = 0; i < m; i++) { int u = prerequisites[i].first, v = prerequisites[i].second; inq[v]++; add_edge(u, v); } queue<int> q; for (int i = 0; i < numCourses; i++) { if (!inq[i]) q.push(i); } while (!q.empty()) { num++; int u = q.front(); q.pop(); for (int i = head[u]; ~i; i = G[i].next) { if (--inq[G[i].to] == 0) q.push(G[i].to); } } delete []G; delete []inq, delete []head; return num == numCourses; } private: int tot, *inq, *head; struct edge { int to, next; }*G; inline void add_edge(int u, int v) { G[tot].to = v, G[tot].next = head[u], head[u] = tot++; } };
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