LeetCode - 207. Course Schedule
207. Course Schedule #
Problem's Link
#
----------------------------------------------------------------------------
Mean:
给定一个有向图,判断是否存在top_sort序列.
analyse:
转换为:判断是否存在环.
若存在环,肯定不能找到top_sort序列.
判环的方式有很多:SPFA,top_sort,BFS,DFS...随便选一种就行.
Time complexity: O(N)
view code
我的代码:
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-03-16-09.31
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);
class Solution
{
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites)
{
unordered_map<int,vector<int>> graph;
build_graph(prerequisites,graph);
unordered_set<int> zero;
vector<int> in_degree(numCourses,0);
count_degree(graph,zero,in_degree);
while(zero.size()>0)
{
int cur=*zero.begin();
zero.erase(cur);
for(auto ptr:graph[cur])
{
--in_degree[ptr];
if(!in_degree[ptr])
zero.insert(ptr);
}
}
for(auto ptr:in_degree)
if(ptr>0) return false;
return true;
}
void build_graph(auto prerequisites,auto& graph)
{
for(auto ptr:prerequisites)
graph[ptr.second].push_back(ptr.first);
}
void count_degree(auto graph,auto& zero,auto& in_degree)
{
for(auto ptr1:graph)
for(auto ptr2:ptr1.second)
++in_degree[ptr2];
for(int i=0;i<in_degree.size();++i)
if(!in_degree[i]) zero.insert(i);
}
};
int main()
{
int num,n;
while(cin>>num>>n)
{
vector<pair<int, int>> prerequisites(n);
for(int i=0;i<n;++i)
cin>>prerequisites[i].first>>prerequisites[i].second;
Solution solution;
bool res=solution.canFinish(num,prerequisites);
if(res) cout<<"Yes."<<endl;
else cout<<"No."<<endl;
}
return 0;
}
/*
*/
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-03-16-09.31
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);
class Solution
{
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites)
{
unordered_map<int,vector<int>> graph;
build_graph(prerequisites,graph);
unordered_set<int> zero;
vector<int> in_degree(numCourses,0);
count_degree(graph,zero,in_degree);
while(zero.size()>0)
{
int cur=*zero.begin();
zero.erase(cur);
for(auto ptr:graph[cur])
{
--in_degree[ptr];
if(!in_degree[ptr])
zero.insert(ptr);
}
}
for(auto ptr:in_degree)
if(ptr>0) return false;
return true;
}
void build_graph(auto prerequisites,auto& graph)
{
for(auto ptr:prerequisites)
graph[ptr.second].push_back(ptr.first);
}
void count_degree(auto graph,auto& zero,auto& in_degree)
{
for(auto ptr1:graph)
for(auto ptr2:ptr1.second)
++in_degree[ptr2];
for(int i=0;i<in_degree.size();++i)
if(!in_degree[i]) zero.insert(i);
}
};
int main()
{
int num,n;
while(cin>>num>>n)
{
vector<pair<int, int>> prerequisites(n);
for(int i=0;i<n;++i)
cin>>prerequisites[i].first>>prerequisites[i].second;
Solution solution;
bool res=solution.canFinish(num,prerequisites);
if(res) cout<<"Yes."<<endl;
else cout<<"No."<<endl;
}
return 0;
}
/*
*/
下面是discuss区的代码,看了一下,感觉还有很多可以优化的地方,我在代码中注释出来了.
代码1:
bool canFinish(int numCourses, vector<vector<int>>& prerequisites)
{
/**< matrix的第二维没必要用unordered_set,因为不管用不用hash都得扫一遍,用hash反而更慢 */
vector<unordered_set<int>> matrix(numCourses); // save this directed graph
for(int i = 0; i < prerequisites.size(); ++ i)
matrix[prerequisites[i][1]].insert(prerequisites[i][0]);
vector<int> d(numCourses, 0); // in-degree
for(int i = 0; i < numCourses; ++ i)
for(auto it = matrix[i].begin(); it != matrix[i].end(); ++ it)
++ d[*it];
for(int j = 0, i; j < numCourses; ++ j)
{
/**< 这儿可以动态维护一个set,存放入度为0的结点,就不必每次都从头开始找了 */
for(i = 0; i < numCourses && d[i] != 0; ++ i); // find a node whose in-degree is 0
if(i == numCourses) // if not find
return false;
d[i] = -1;
for(auto it = matrix[i].begin(); it != matrix[i].end(); ++ it)
-- d[*it];
}
return true;
}
{
/**< matrix的第二维没必要用unordered_set,因为不管用不用hash都得扫一遍,用hash反而更慢 */
vector<unordered_set<int>> matrix(numCourses); // save this directed graph
for(int i = 0; i < prerequisites.size(); ++ i)
matrix[prerequisites[i][1]].insert(prerequisites[i][0]);
vector<int> d(numCourses, 0); // in-degree
for(int i = 0; i < numCourses; ++ i)
for(auto it = matrix[i].begin(); it != matrix[i].end(); ++ it)
++ d[*it];
for(int j = 0, i; j < numCourses; ++ j)
{
/**< 这儿可以动态维护一个set,存放入度为0的结点,就不必每次都从头开始找了 */
for(i = 0; i < numCourses && d[i] != 0; ++ i); // find a node whose in-degree is 0
if(i == numCourses) // if not find
return false;
d[i] = -1;
for(auto it = matrix[i].begin(); it != matrix[i].end(); ++ it)
-- d[*it];
}
return true;
}
代码2:
class Solution
{
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites)
{
/**< matrix的第二维没必要用unordered_set,因为不管用不用hash都得扫一遍,用hash反而更慢 */
vector<unordered_set<int>> graph = make_graph(numCourses, prerequisites);
vector<int> degrees = compute_indegree(graph);
for (int i = 0; i < numCourses; i++)
{
int j = 0;
/**< 这儿可以动态维护一个set,存放入度为0的结点,就不必每次都从头开始找了 */
for (; j < numCourses; j++)
if (!degrees[j]) break;
if (j == numCourses) return false;
degrees[j] = -1;
for (int neigh : graph[j])
degrees[neigh]--;
}
return true;
}
private:
vector<unordered_set<int>> make_graph(int numCourses, vector<pair<int, int>>& prerequisites)
{
vector<unordered_set<int>> graph(numCourses);
for (auto pre : prerequisites)
graph[pre.second].insert(pre.first);
return graph;
}
vector<int> compute_indegree(vector<unordered_set<int>>& graph)
{
vector<int> degrees(graph.size(), 0);
for (auto neighbors : graph)
for (int neigh : neighbors)
degrees[neigh]++;
return degrees;
}
};
{
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites)
{
/**< matrix的第二维没必要用unordered_set,因为不管用不用hash都得扫一遍,用hash反而更慢 */
vector<unordered_set<int>> graph = make_graph(numCourses, prerequisites);
vector<int> degrees = compute_indegree(graph);
for (int i = 0; i < numCourses; i++)
{
int j = 0;
/**< 这儿可以动态维护一个set,存放入度为0的结点,就不必每次都从头开始找了 */
for (; j < numCourses; j++)
if (!degrees[j]) break;
if (j == numCourses) return false;
degrees[j] = -1;
for (int neigh : graph[j])
degrees[neigh]--;
}
return true;
}
private:
vector<unordered_set<int>> make_graph(int numCourses, vector<pair<int, int>>& prerequisites)
{
vector<unordered_set<int>> graph(numCourses);
for (auto pre : prerequisites)
graph[pre.second].insert(pre.first);
return graph;
}
vector<int> compute_indegree(vector<unordered_set<int>>& graph)
{
vector<int> degrees(graph.size(), 0);
for (auto neighbors : graph)
for (int neigh : neighbors)
degrees[neigh]++;
return degrees;
}
};
作者:北岛知寒
出处:https://www.cnblogs.com/crazyacking/p/5283462.html
版权:本作品采用「署名-非商业性使用-相同方式共享 4.0 国际」许可协议进行许可。
分类:
ACM/LC
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