图的着色和会场分配问题是一样的

先贴图的着色问题代码，回溯算法介绍见：https://blog.csdn.net/qq_34793133/article/details/80685985

问题的设计思想我会及时更新

#include <iostream>
using namespace std;

const int N = 50;
int color[N] = { 0 };
int trace[N] = { 0 };
int graph[N][N] = { 0 };
int PointNum; int methods, color_limit;

void Initiate()
{
	cout << "请输入点个数： " << endl;
	cin >> PointNum;

	cout << "请输入颜色数" << endl;
	cin >> color_limit;
	cout << "请您依据图中节点的邻接关系输入邻接矩阵graph[][] " << endl;
	for (int i = 1; i <= PointNum; ++i)
		for (int j = 1; j <= PointNum; ++j)
		{
			cin >> graph[i][j];
		}

}


//判断该点着此种颜色是否与邻点重复
bool OK(int i)
{
	for (int m = 1; m <= PointNum; ++m)
	{
		if ((graph[i][m]==1)&&(color[m]== color[i]))
			return false;

	}
	return true;
}

void Traceback(int i)
{
	if (i > PointNum)
	{
		methods++; 
		//return 1;
		
	}
	else
	{

		for (int k = 1; k <= color_limit; ++k)
		{
			color[i] = k;
			if (OK(i))
			{
				Traceback(i + 1);
			}
			color[i] = 0;//color[i] = 0;必须有，若去掉，当遇到某点对所有颜色都不合适的时候，该点将保留颜色编号最大的颜色color[color_limit]
			//这样的话再回溯i-1的节点是不对的，对于i之前的节点比如i-1，color[i]应为0
		}
	}
}

int main()
{
	Initiate();
	Traceback(1);
	cout << "一共" << methods << "种" << "着色方法" << endl;

    return 0;
}

参考：https://blog.csdn.net/jeffleo/article/details/54586046

转载需要注明出处：https://blog.csdn.net/qq_34793133/article/details/80696862