LeetCode Course Schedule BFS

思路：

BFS + 拓扑排序

Status：MLE

Reason：用二维数组

vector<vector<int>>

存图信息，有太多没用的信息0，占用了很大空间。

看了DISCUSS后的优化：

用

vector<unordered_set<int>>

存图信息，表示为每个节点对应连接到哪些节点。

class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<int> inDegree(numCourses, 0);
vector<vector<int>> graph(numCourses, vector<int>(numCourses, 0));
queue<int> q;
for(int i = 0; i < numCourses; ++i) {
if(inDegree[i] == 0) q.push(i);
}
while(!q.empty()) {
int start_node_id = q.front();
q.pop();
for(int i = 0; i < numCourses; ++i) {
if(graph[start_node_id][i] == 1) {
inDegree[i]--;
if(inDegree[i] == 0) {
q.push(i);
}
}
}
}
for(int i = 0; i < numCourses; ++i) {
if(inDegree[i] != 0) {
return false;
}
}
return true;
}
};

Staus：AC

class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<unordered_set<int>> need(numCourses);
for (size_t i = 0; i != prerequisites.size(); ++i)
need[prerequisites[i].second].insert(prerequisites[i].first);
vector<int> indegree(numCourses);
for (int i = 0; i != numCourses; ++i)
for(auto it = need[i].begin(); it != need[i].end(); ++it)
++indegree[*it];
queue<int> zeros;
for (int i = 0; i != numCourses; ++i)
if (indegree[i] == 0)
zeros.push(i);
while (!zeros.empty()) {
int seq = zeros.front();
zeros.pop();
for (auto it = need[seq].begin(); it != need[seq].end(); ++it)
if (--indegree[*it] == 0)
zeros.push(*it);
--numCourses;
}
return numCourses == 0;
}
};