（POJ 1258）Prim算法 最小生成树

Prim 只于顶点有关
基本思想

1.清空生成树，任取一个顶点加入生成树

2.在那些一个端点在生成树里，另一个端点不在生成树里的边中，选取一条权最小的边，将它和另一个端点加进生成树

3.重复步骤2，直到所有的顶点都进入了生成树为止，此时的生成树就是最小生成树

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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

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AC代码：

#include <iostream>
#include <algorithm>
#include  <cstdio>
#include <cstring>
using namespace std;
#define MAX 0x3f3f3f3f
int map[101][101];
int n;
void prim()
{
    int i,j,now;
    int sum=0;
    int logo[101];  //标记数组 标记某点是否已经加入最小生成树集合 也可以为bool类型
    int dis[101];    //记录邻接点的最短边
    for(i=1;i<n;i++)
    {
        dis[i]=map[0][i];
        logo[i]=0;
    }
    logo[0]=1;  //将起始点加入最小生成树集合
    for(i=1;i<=n-1;i++) //找生成树集合点集相连最小权值的边 
    {
        now=MAX;
        int min=MAX;
        for(j=1;j<n;j++)
        {
            if(logo[j]==0&&dis[j]<min)
            {
                now=j;
                min=dis[j];
            }
        }
        logo[now]=1;  //加入最小生成树集合
        sum+=min;  //记录权值和
        for(j=1;j<n;j++)    //更新dis数组
        {
            if(logo[j]==0&&dis[j]>map[now][j])
                dis[j]=map[now][j];
        }
    }
    printf("%d\n",sum);
}
int main()
{
    int i,j;
    while(scanf("%d",&n)!=EOF&&n)
    {
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
                scanf("%d",&map[i][j]);  //输入边的信息
        prim();
    }
    return 0;
}