题目连接
https://leetcode.com/problems/course-schedule/
Course Schedule
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++;
}
};