# 如何使用回溯找到图着色的时间复杂度？

##### 问题描述投票：3回答：1

``````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");
}
``````
algorithm graph colors backtracking
##### 1个回答
0

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