普里姆算法求最小生成树

#include "iostream"
using namespace std;

const int num = 9; //节点个数
#define Infinity 65535;
//本例中以节点0作为生成树的起始节点
void MinSpanTree_Prime(int graphic[num][num]){
	int lowcost[num]; //记录从节点num到生成树的最短距离，如果为0则表示该节点已在生成树中
	int adjvex[num]; //记录相关节点，如:adjvex[1] = 5表示最小生成树中节点1和5有路径相连
	int sum = 0; // 记录最小生成树边权重之和
	memset(adjvex, 0, sizeof(adjvex));
	//选取0节点作为生成树的起点
	for (int i = 0; i < num; i++)
		lowcost[i] = graphic[0][i];

	for (int i = 1; i < num; i++){
		int min = Infinity;
		int index;
		for (int j = 1; j < num; j++){
			if (lowcost[j] != 0 && lowcost[j] < min){
				index = j;
				min = lowcost[j];
			}
		}
		sum += min;
		lowcost[index] = 0; //将当前节点放入生成树中
		cout << adjvex[index] << " -> " << index << endl;
		//修正其他节点到生成树的最短距离
		for (int j = 1; j < num; j++){
			if (lowcost[j] != 0 && graphic[index][j] < lowcost[j]){
				lowcost[j] = graphic[index][j];
				adjvex[j] = index;
			}
		}
	}
	cout << "sum = " << sum << endl;
}

int main(){
	int graphic[num][num];
	for (int i = 0; i < num; i++)
	for (int j = 0; j < num; j++){
		if (i == j)
			graphic[i][j] = 0;
		else
			graphic[i][j] = Infinity;
	}
	graphic[0][1] = 1;
	graphic[0][2] = 5;
	graphic[1][0] = 1;
	graphic[1][2] = 3;
	graphic[1][3] = 7;
	graphic[1][4] = 5;
	graphic[2][0] = 5;
	graphic[2][1] = 3;
	graphic[2][4] = 1;
	graphic[2][5] = 7;
	graphic[3][1] = 7;
	graphic[3][4] = 2;
	graphic[3][6] = 3;
	graphic[4][1] = 5;
	graphic[4][2] = 1;
	graphic[4][3] = 2;
	graphic[4][5] = 3;
	graphic[4][6] = 6;
	graphic[4][7] = 9;
	graphic[5][2] = 7;
	graphic[5][4] = 3;
	graphic[5][7] = 5;
	graphic[6][3] = 3;
	graphic[6][4] = 6;
	graphic[6][7] = 2;
	graphic[6][8] = 7;
	graphic[7][4] = 9;
	graphic[7][5] = 5;
	graphic[7][6] = 2;
	graphic[7][8] = 4;
	graphic[8][6] = 7;
	graphic[8][7] = 4;

	MinSpanTree_Prime(graphic);

	return 0;
}