最小生成树

一个有 n 个结点的连通图的生成树是原图的极小连通子图，且包含原图中的所有 n 个结点，并且有保持图连通的最少的边。

今天看了普利姆算法.记录一下.

<pre name="code" class="cpp">// bogo.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <stdio.h>#include <iostream>using namespace std;#define MAX_PATH 65535int date[9][9] = {{0,10,MAX_PATH,MAX_PATH,MAX_PATH,11,MAX_PATH,MAX_PATH,MAX_PATH},{10,0,18,MAX_PATH,MAX_PATH,MAX_PATH,16,MAX_PATH,12},{MAX_PATH,MAX_PATH,0,22,MAX_PATH,11,MAX_PATH,MAX_PATH,8},{MAX_PATH,MAX_PATH,22,0,20,MAX_PATH,MAX_PATH,16,21},{MAX_PATH,MAX_PATH,MAX_PATH,20,0,26,MAX_PATH,7,MAX_PATH},{11,MAX_PATH,MAX_PATH,MAX_PATH,26,0,17,MAX_PATH,MAX_PATH},{MAX_PATH,16,MAX_PATH,MAX_PATH,MAX_PATH,17,0,19,MAX_PATH},{MAX_PATH,MAX_PATH,MAX_PATH,16,7,MAX_PATH,19,0,MAX_PATH},{MAX_PATH,12,8,21,MAX_PATH,MAX_PATH,MAX_PATH,MAX_PATH,0}};int _tmain(int argc, _TCHAR* argv[]){int lowest[9];//这个数组可以理解为下标为i的点,与现在已经加入生成树各点的最小权值.int adjvex[9];//a[i] = k 就是表示i点与树中的k点之间权值最小.如果目前树种没有与i相连,则a[i] = 0int min;int j,k;int length = 9;lowest[0] = 0;adjvex[0] = 0;int i;//初始化for (i = 1; i < length; i++){adjvex[i] = 0;lowest[i] = date[0][i];}for (i = 1; i < length; i++){//初始化最小权值为MAXmin = MAX_PATH;j = 1;k = 0;while (j < length){if (lowest[j] != 0&&lowest[j] < min){min = lowest[j];k = j;}j++;}printf("%d %d\n",adjvex[k],min);lowest[k] = 0;//表示此点已经加入了最小生成树for (j = 1; j < length; j++){if (lowest[j] != 0&&date[k][j] < lowest[j]){lowest[j] = date[k][j];adjvex[j] = k;}}}return 0;}