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法律研究使用DFS检查图是否是二分的C++程序
二分图是一种可以用两种颜色进行图着色的图，例如，某一组中的顶点使用同一种颜色进行着色。这是一个使用DFS检查图是否是二分的C++程序。
算法
Begin 1. An array color[] is used to stores 0 or 1 for every node which denotes opposite colors. 2. Call function DFS from any node. 3. If the node w has not been visited previously, then assign ! color[v] to color[w] and call DFS again to visit nodes connected to w. 4. If at any instance, color[u] is equal to !color[v], then the node is bipartite. 5. Modify the DFS function End
示例
#include<iostream>
#include <bits/stdc++.h>
using namespace std;
void addEd(vector<int> adj[], int w, int v) //adding edge to the graph {
adj[w].push_back(v); //add v to w’s list
adj[v].push_back(w); //add w to v’s list
}
bool Bipartite(vector<int> adj[], int v,
vector<bool>& visited, vector<int>& color) {
for (int w : adj[v]) {
// if vertex w is not explored before
if (visited[w] == false) {
// mark present vertex as visited
visited[w] = true;
color[w] = !color[v]; //mark color opposite to its parents
if (!Bipartite(adj, w, visited, color))
return false;
}
// if two adjacent are colored with same color then the graph is not bipartite
else if (color[w] == color[v])
return false;
}
return true;
}
int main() {
int M = 6;
vector<int> adj[M + 1];
// to keep a check on whether
// a node is discovered or not
vector<bool> visited(M + 1);
vector<int> color(M + 1); //to color the vertices of the graph with 2 color
addEd(adj, 3,2);
addEd(adj, 1,4 );
addEd(adj, 2, 1);
addEd(adj, 5,3);
addEd(adj, 6,2);
addEd(adj, 3,1);
visited[1] = true;
color[1] = 0;
if (Bipartite(adj, 1, visited, color)) {
cout << "Graph is Bipartite";
} else {
cout << "Graph is not Bipartite";
}
return 0;
}输出
Graph is not Bipartite
更新于： 30-Jul-2019
207 次浏览
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