poj 1258 Agri

Agri-Net Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 44487   Accepted: 18173

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

題意： 有n個農場，已知這n個農場都互相相通，有一定的距離，現在每個農場需要裝光纖，問怎麼安裝光纖能將所有農場都連通起來，並且要使光纖距離最小，輸出安裝光纖的總距離。
思路： 最小生成樹，給出的二維矩陣代表他們的距離，prim算法求解即可。
代碼：

#include
#include
#include
using namespace std;
#define maxn 110
#define inf 1<<29

int Map[maxn][maxn],vis[maxn],low[maxn];
int n;
//第一種
void prim()
{
    int Min,pos,sum=0;
    for(int i=1;i<=n;i++)
    {
        low[i]=Map[1][i];
        vis[i]=0;
    }
    for(int i=1;i<=n;i++)
    {
        Min=inf;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j] && low[j]Map[pos][j])
                low[j]=Map[pos][j];
        }
    }
    for(int i=1;i<=n;i++)
        sum+=low[i];
    printf(%d
,sum);
}
/*第二種
void prim()
{
    int Min,pos,sum=0;
    memset(vis,0,sizeof(vis));
    vis[1]=1,pos=1;
    for(int i=1;i<=n;i++)
    {
        if(i!=pos)
            low[i]=Map[pos][i];
    }
    for(int i=1;iMap[pos][j])
                low[j]=Map[pos][j];
        }
    }
    printf(%d
,sum);
}
*/
int main()
{
    while(~scanf(%d,&n))
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                scanf(%d,&Map[i][j]);
        prim();
    }
    return 0;
}