Course Schedule

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.


图论中的一个典型题目，求有向图的拓扑排序，这里相对简化了一些，只需要判断一个图是否是DAG。

思路：

所有的拓扑排序都从一个入度为0的点开始（本例中就是一个不需要任何铺垫课程的课程），从图中移除该点和从它发出的所有边。重复这一过程直到所有的点都移除（表明该图是一个DAG）或者剩余的点里没有入度为0的点了（不是DAG）。

代码：

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int> >& prerequisites) {
        if(numCourses == 0 || prerequisites.empty()) return true;
        int nedge = prerequisites.size();

        vector<int> indeg(numCourses, 0);
        vector<vector<int> > edges(numCourses, vector<int>());
        vector<bool> visited(numCourses, false);
        int count = 0;

        for(int i = 0; i < nedge; ++i){
            ++indeg[prerequisites[i].first];
            edges[prerequisites[i].second].push_back(prerequisites[i].first);
        }

        while(count < numCourses){
            int i = 0;
            for(; i < numCourses; ++i) if(!visited[i] && indeg[i] == 0) break;
            if(i == numCourses) return false;
            visited[i] = true;
            int nout = edges[i].size();
            for(int j = 0; j < nout; ++j) --indeg[edges[i][j]];
            ++count;
        }
        return true;
    }
};