207. Course Schedule

class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
map<int,int> count;
for(int i=0;i<numCourses;i++){
count[i]=0;
}

for(int i=0;i<prerequisites.size();i++){
count[prerequisites[i].first]++;
}
queue<int> que;
for(int i=0;i<count.size();i++){
if(count[i]==0){
que.push(i);
}
}
int total=que.size();

while(!que.empty()){
int cur=que.front();

que.pop();
for(int i=0;i<prerequisites.size();i++){
if(prerequisites[i].second==cur){
count[prerequisites[i].first]--;
if(count[prerequisites[i].first]==0){
que.push(prerequisites[i].first);
total++;
}
}
}
}
return total==numCourses;
}
};

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.

分析：先初始化一个map放置每个节点的入度，将每个入度为0的节点加入队列，对于队列中的每个元素，将其能到达的节点的入度减一，若此时能够得到入度为0的节点，继续将节点推进队列。计算队列一共被推进了多少个元素，若等于课程总数，则输出true，反之输出false。