Week4(2)
上周犯了个小错误，在写第3周的博客时，不小心直接在第2周的博客上改了，然后第2周的博客直接被第3周的覆盖了，所以这周写2篇补回来。
Graph
question source: Is Graph Bipartite?
question description
Given an undirected graph, return true if and only if it is bipartite.
Recall that a graph is bipartite if we can split it’s set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.
The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists. Each node is an integer between 0 and graph.length - 1. There are no self edges or parallel edges: graph[i] does not contain i, and it doesn’t contain any element twice.
Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation:
The graph looks like this:
0----1
| |
| |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.
Example 2:
Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation:
The graph looks like this:
0----1
| \ |
| \ |
3----2
We cannot find a way to divide the set of nodes into two independent subsets.
Note:
- graph will have length in range [1, 100].
- graph[i] will contain integers in range [0, graph.length - 1].
- graph[i] will not contain i or duplicate values.
- The graph is undirected: if any element j is in graph[i], then i will be in graph[j].
这周学习了图的分解，选这道题是因为老师上课讲课后习题的时候讲了类似的题的算法，刚好就看到有，就实现了。在中文教材的111页，3.7题，二部图是这样的图，其顶点可以划分为两人子集，并且子集内部的顶点之间没有边相连。这条性质可以有多种表述方式，例如，无向图是二部图，当且仅当用两种颜色就可以为它着色，无向图是二部图当且仅当它不包含长度为奇数的环。 这几种表示都是可以的，用颜色好理解一点。遍历这个图，用两种颜色来标记点，如果相临的两个点颜色相同，则不是二部图。
解决方法
做法1，使用BFS，每一层的颜色都和上一层的颜色不同，上色之后检查顶点有没有指向相同颜色的顶点。具体实现方法是，用一个color的数组，为0时表示还没被染色，也就是说没有被访问过，用1和-1来区别不同的颜色，这样对下一层的点染色时，只要将这一层的颜色取反就行了。
实现代码如下
class Solution {
public:
bool isBipartite(vector<vector<int>>& graph) {
//bfs
int size = graph.size();
int n = 0;
queue<int> q;
int color[size];
memset(color, 0, sizeof(color));
q.push(0);
color[0] = 1;
while(n != size){
int v = q.front();
n++;
for(int i = 0; i < graph[v].size(); i++){
if(color[graph[v][i]] == color[v]){
return false;
}else if(color[graph[v][i]] == 0){
color[graph[v][i]] = (-1) * color[v];
q.push(graph[v][i]);
}
}
q.pop();
// The graph is not connected
if(q.size() == 0 && n != size){
for(int i = 0; i < size; i++){
if(color[i] == 0){
q.push(i);
color[i] = 1;
break;
}
}
}
}
return true;
}
};
第2种方法，做法1，使用DFS，每个点的相临点都染上不同的颜色，同时对正在染色的点进行检查，如果发现相临的点有相同颜色，则不是二部图。DFS用递归实现。
实现代码如下
class Solution {
public:
bool isBipartite(vector<vector<int>>& graph) {
//dfs
int size = graph.size();
int color[size];
memset(color, 0, sizeof(color));
int n = 0;
int v = 0;
bool flag = true;
while(true){
color[v] = 1;
visit(v, graph, n, color, flag);
if(!flag){
return false;
}
if(n == size){
return true;
}
for(int i = 0; i < graph.size(); i++){
if(color[i] == 0){
v = i;
break;
}
}
}
}
void visit(int v, vector<vector<int>> & graph, int& n, int* color, bool& flag){
if(!flag){
return;
}
n++;
for(int i = 0; i < graph[v].size(); i++){
if(color[graph[v][i]] == color[v]){
flag = false;
return;
}else if(color[graph[v][i]] == 0){
color[graph[v][i]] = -1 * color[v];
visit(graph[v][i], graph, n, color, flag);
}
}
}
};
思路是很清晰的，但实现起来就有各种问题。一开始时用一个数组来表示是否访问过，用另一个数组来记录颜色。这样实现起来即不方便，而且也没必要。后来觉得用一个数组就能实现这个功能了。