Agri-Net（最小生成树）

Agri-Net

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 54170		Accepted: 22491

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

代码如下：

#include<stdio.h>
int inf=99999999;
int main()
{
    int n,i,j,min,map[111][111];//存所有距离
    int book[111];//标记是否走过
    int dis[111];
    while(~scanf("%d",&n))
    {
        for(i=0; i<n; i++) //初始化，全部清零
        {
            dis[i]=0;
            book[i]=0;
        }
        int count=0;//生成树中节点的个数
        int sum=0;
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
                scanf("%d",&map[i][j]);
        for(i=0; i<n; i++)//存入0到其他点的距离
            dis[i]=map[0][i];
        book[0]=1;//标记0点已经走过
        count++;
        while(count<n)
        {
            min=inf;
            for(i=0; i<n; i++)//遍历查找到0点的最短路径
                if(dis[i]<min&&book[i]==0)
                {
                    min=dis[i];
                    j=i;
                }
            book[j]=1;
            count++;
            sum+=dis[j];
            /*每次将一个节点接到生成树之后，将整一棵树看作是一个点（0点）
            **路径更新只需要更新到0点的距离
            */
            for(i=0; i<n; i++)//更新到0点的最短路径
                if(book[i]==0&&dis[i]>map[j][i])
                    dis[i]=map[j][i];

        }
        printf("%d\n" ,sum);
    }
    return 0;
}

详细讲解：

《啊哈！算法》 P219