如何使用回溯找到图着色的时间复杂度?

时间:2018-04-17 20:55:02

标签: algorithm graph colors backtracking

我必须使用回溯来找出图着色问题的时间复杂度。我发现某处是O(n * m ^ n),其中n =无顶点,m =颜色数。

假设我的代码在下面给出了如何找到时间复杂度?

bool isSafe (int v, bool graph[V][V], int color[], int c)
{
    for (int i = 0; i < V; i++)
        if (graph[v][i] && c == color[i])
            return false;
    return true;
}

bool graphColoringUtil(bool graph[V][V], int m, int color[], int v)
{
    if (v == V)
        return true;

    for (int c = 1; c <= m; c++)
    {
        if (isSafe(v, graph, color, c))
        {
           color[v] = c;

           if (graphColoringUtil (graph, m, color, v+1) == true)
             return true;

           color[v] = 0;
        }
    }

    return false;
}
bool graphColoring(bool graph[V][V], int m)
{
    int *color = new int[V];
    for (int i = 0; i < V; i++)
       color[i] = 0;

    if (graphColoringUtil(graph, m, color, 0) == false)
    {
      printf("Solution does not exist\n");
      return false;
    }

    printSolution(color);
    return true;
}
void printSolution(int color[])
{
    printf("Solution Exists:"
            " Following are the assigned colors \n");
    for (int i = 0; i < V; i++)
      printf(" %d ", color[i]);
    printf("\n");
} 

1 个答案:

答案 0 :(得分:1)

graphutil方法自身将执行n次。它在c循环中,并且c升至m。 现在c循环由于递归(即m ^ n)进行了n次,而递归进行了n次,所以总计为O(nm ^ n)