- Find the Weak Connected Component in the Directed Graph
Find the number Weak Connected Component in the directed graph. Each node in the graph contains a label and a list of its neighbors. (a weak connected component of a directed graph is a maximum subgraph in which any two vertices are connected by direct edge path.)
Example
Example 1:
Input: {1,2,4#2,4#3,5#4#5#6,5}
Output: [[1,2,4],[3,5,6]]
Explanation:
1----->2 3-->5
\ | ^
\ | |
\ | 6
\ v
->4
Example 2:
Input: {1,2#2,3#3,1}
Output: [[1,2,3]]
Clarification
graph model explaination:
https://www.lintcode.com/help/graph
Notice
Sort the elements of a component in ascending order.
解法1:Union-Find
这题最好用并查集做。用BFS/DFS也可以做,但因为是有向图,必须把有向链接转换成无向链接才行。为什么呢?
比如说Input: {1,2,4#2,4#3,5#4#5#6,5}
Output: [[1,2,4],[3,5,6]]
Explanation:
1----->2 3-->5
\ | ^
\ | |
\ | 6
\ v
->4
如果用BFS/DFS的话,到了3就找不到6了,所以没法知道3和6是一组。
注意:
1)merge()的写法要注意。特别是最后是father[fatherX]=fatherY。不能写成father[x]=fatherY,这样x是连到了fatherY,但是fatherX到fatherY的链接就断了。为了简单起见,可以将merge(int x, int y)写成这样的模板:
void merge (int x, int y) {
int x = find(x);
int y = find(y);
if (x != y) father[x]=y;
}
-
当union完成后,以input={1,2,3,4#2,3#3#4}为例,其各个节点的father如下。
1’s father is 3
2’s father is 4
3’s father is 4
4’s father is 4
此时并不能看出它们都是同一个father,所以最后的for循环里面还是必须用find(x)来找到各个节点的最上层祖宗,不能直接用father[x]。 -
father在这里必须用map,不能直接用数组,因为label可能很大(不一定是1,2,3,4,5),另外也可能是负数。
代码如下:
/**
- Definition for Directed graph.
- struct DirectedGraphNode {
-
int label; -
vector<DirectedGraphNode *> neighbors; -
DirectedGraphNode(int x) : label(x) {}; - };
*/
class Solution {
public:
/*
* @param nodes: a array of Directed graph node
* @return: a connected set of a directed graph
/
vector<vector> connectedSet2(vector<DirectedGraphNode> nodes) {
int n = nodes.size();
//initialize, each node's father is itself
for (auto node : nodes) {
father[node->label] = node->label;
}
//union
for (int i = 0; i < n; ++i) {
for (auto neighbor : nodes[i]->neighbors) {
merge(nodes[i]->label, neighbor->label);
}
}
map<int, vector<int>> component; //father, children
for (int i = 0; i < n; ++i) {
int findFather = find(father[nodes[i]->label]);
if (component.find(findFather) == component.end()) {
component[findFather] = vector<int>();
}
component[findFather].push_back(nodes[i]->label);
}
for (auto m : component) {
result.push_back(m.second);
}
return result;
}
private:
map<int, int> father;
//vector father; //wrong, because label can be negative
vector<vector> result;
int find(int x) {
if (x == father[x]) {
return x;
}
father[x] = find(father[x]);
return father[x];
}
void merge(int x, int y) {
int fatherX = find(x); //not fatherX = father[x];
int fatherY = find(y);
if (fatherX != fatherY) {
father[fatherX] = fatherY; //not father[y]
}
}
};
本文探讨了如何在有向图中寻找弱连通分量的方法,主要介绍了使用并查集算法进行高效求解的过程。通过具体实例说明了并查集在处理此类问题上的优势,并提供了详细的代码实现。
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