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法律研究使用 BFS 檢查圖是否是二分的 C++ 程式
二分圖是如果圖著色可以只用兩種顏色進行（即一組中所有頂點都用同一種顏色著色），則二分圖就是一幅圖。本文提供了一個使用 BFS 檢查圖是否為二分圖的 C++ 程式。
演算法
Begin Function Bipartite(): 1) Assign a color to the source vertex 2) Color all the neighbors with another color except first one color. 3) Color all neighbor’s neighbor with First color. 4) Like this way, assign color to all vertices such that it satisfies all the constraints of k way coloring problem where k = 2. 5) While assigning colors, if we find a neighbor which is colored with same color as current vertex, then the graph cannot be colored with 2 vertices i.e.; graph is not Bipartite End
示例
#include <iostream>
#include <queue>
#define V 5
using namespace std;
bool Bipartite(int G[][V], int s) {
int colorA[V];
for (int i = 0; i < V; ++i)
colorA[i] = -1;
colorA[s] = 1; //Assign a color to the source vertex
queue <int> q; //Create a queue of vertex numbers and enqueue source vertex for BFS traversal
q.push(s);
while (!q.empty()) {
int w = q.front(); //dequeue a vertex
q.pop();
for (int v = 0; v < V; ++v) //Find all non-colored adjacent vertices {
if (G[w][v] && colorA[v] == -1) //An edge from w to v exists and destination v is not colored {
colorA[v] = 1 - colorA[w]; //Assign alternate color to this adjacent v of w
q.push(v);
} else if (G[w][v] && colorA[v] == colorA[w]) //An edge from w to v exists and destination
//v is colored with same color as u
return false;
}
}
return true; //if all adjacent vertices can be colored with alternate color
}
int main() {
int G[][V] = {{ 0, 1, 0, 0},
{ 1, 0, 0, 0},
{ 0, 0, 0, 1},
{ 1, 0, 1, 0}};
if (Bipartite(G, 0))
cout << "The Graph is Bipartite"<<endl;
else
cout << "The Graph is Not Bipartite"<<endl;
return 0;
}輸出
The Graph is Bipartite
更新於: 2019 年 7 月 30 日
645 次瀏覽
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