poj 1258 Agri-Net（最小生成树果题）

题目链接：http://poj.org/problem?id=1258

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

Source

USACO 102

代码如下：

#include <stdio.h>
#include <string.h>
#define INF 0x3f3f3f3f
#define MAXN 517
//创建m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 m[MAXN][MAXN], low[MAXN], visited[MAXN];
int n;
int prim( )
{
    int i, j;
    int pos, minn, result=0;
    memset(visited,0,sizeof(visited));
    visited[1] = 1;
    pos = 1;          //从某点开始，分别标记和记录该点
    for(i = 1; i <= n; i++)     //第一次给low数组赋值
        if(i != pos)
            low[i] = m[pos][i];
        else
            low[i] = 0;
    for(i = 1; i < n; i++) //再运行n-1次
    {
        minn = INF;   //找出最小权值并记录位置
        for(j=1; j<=n; j++)
        {
            if(visited[j]==0 && minn>low[j])
            {
                minn = low[j];
                pos = j;
            }
        }
        result += minn;   //最小权值累加
        visited[pos] = 1;   //标记该点
        for(j = 1; j <= n; j++)   //更新权值
            if(visited[j]==0 && low[j]>m[pos][j])
                low[j] = m[pos][j];
    }
    return result;
}
int main()
{
    int i,j,ans;
    while(scanf("%d",&n)!=EOF)
    {
        memset(m,INF,sizeof(m));   //所有权值初始化为最大
        for(i = 1; i <= n; i++)
            for(j = 1; j <= n; j++)
            {
                scanf("%d",&m[i][j]);
            }
        ans=prim( );
        printf("%d\n",ans);
    }
    return 0;
}

poj 1258 Agri-Net（最小生成树果题）