poj1258Agri-Net

最新推荐文章于 2021-07-27 09:17:43 发布

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本文介绍了一种使用Prim算法解决最小生成树问题的方法，旨在帮助读者理解如何通过该算法来寻找连接所有节点所需的最短边集合。文章提供了一个具体的编程实例，包括完整的C++代码实现。

Agri-Net
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 55803 Accepted: 23143
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

求用最少的代价链接所有的建筑

最小生成数prim算法

#include <cstdio>
#include <iostream>
#include <cstring>
#define INF 0x3f3f3f
using namespace std;
#define N 110
int map[N][N];
int n;
int vist[N];
int num[N];
struct node
{
    int u,v,w;
};

int prim()
{
    int i,j,min,pos,result=0;
    memset(vist,0,sizeof(vist));
    vist[1]=1;
    pos=1;
    for(i=1; i<=n; i++)
    {
        if(i!=pos)
        {
            num[i]=map[pos][i];
        }
    }
    for(i=1; i<n; i++)
    {
        min=INF;
        for(j=1; j<=n; j++)
        {
            if(!vist[j]&&min>num[j])
            {
                min=num[j];
                pos=j;
            }
        }

        result+=min;
        vist[pos]=1;
        for(j=1; j<=n; j++)
        {
            if(!vist[j]&&num[j]>map[pos][j])
            {
                num[j]=map[pos][j];
            }
        }
    }
    return result;
}
int main()
{
    int i,v,j,ans;
    while(scanf("%d",&n)!=EOF)
    {
        memset(map,INF,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\n",ans);
    }
    return 0;
}