最小生成树之prim的算法实现

今天研究小最小生成树的实现。所谓生成树，就是n个点之间连成n-1条边的图形。而最小生成树，就是权值（两点间直线的值）之和的最小值。

用 POJ 1258 Agri-Net 为例演示

Description

Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.
Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.
Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.
The distance between any two farms will not exceed 100,000.

Input

The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.

Output

For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.

Sample Input

4
0 4 9 21
4 0 8 17
9 8 0 16
21 17 16 0

Sample Output

28

题意：简单介绍一下题意。农民要建立互联网络，目的使村庄里所有的农民连上网，并且总费用最小。多组数据，每组数据给出一个n，然后给出n * n大小的无向图的邻接矩阵表示，值表示边权。要求输出最小生成树的权值和。

大致如图：

代码如下：

#include <stdio.h>
#include <string.h>
#define MaxInt 0x3f3f3f3f
#define N 110
//创建map二维数组储存图表，low数组记录每2个点间最小权值，visited数组标记某点是否已访问
int map[N][N],low[N],visited[N];
int n;
 
int prim()
{
    int i,j,pos,min,result=0;
    memset(visited,0,sizeof(visited));
//从某点开始，分别标记和记录该点
    visited[1]=1;pos=1;
//第一次给low数组赋值
    for(i=1;i<=n;i++)
        if(i!=pos) low[i]=map[pos][i];
//再运行n-1次
    for(i=1;i<n;i++)
    {
//找出最小权值并记录位置
     min=MaxInt;
     for(j=1;j<=n;j++)
         if(visited[j]==0&&min>low[j])
         {
             min=low[j];pos=j;
         }
//最小权值累加
    result+=min;
//标记该点
    visited[pos]=1;
//更新权值
    for(j=1;j<=n;j++)
        if(visited[j]==0&&low[j]>map[pos][j])
            low[j]=map[pos][j];
    }
    return result;
}
 
int main()
{
    int i,v,j,ans;
    while(scanf("%d",&n)!=EOF)
    {
//所有权值初始化为最大
        memset(map,MaxInt,sizeof(map));
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            {
                scanf("%d",&v);
                map[i][j]=map[i][j]=v;
            }
            ans=prim();
            printf("%d
",ans);
    }
    return 0;
}