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;elsegraphic[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;}