poj_1258 Agri-Net

最新推荐文章于 2021-07-28 09:52:27 发布

最新推荐文章于 2021-07-28 09:52:27 发布 · 71 阅读

本文详细介绍了如何解决最小光纤铺设问题，即使用Prim算法来连接一组农场以实现互联网连接。通过实例分析，解释了输入格式、输出要求以及解题思路，并提供了C++代码实现。

Agri-Net

Time Limit: 1000MS	Memory Limit: 10000K
Total Submissions: 28121	Accepted: 11149

题目链接：http://poj.org/problem?id=1258

Description

Farmer John has been electedmayor of his town! One of his campaign promises was to bring internetconnectivity 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 sharehis connectivity with the other farmers. To minimize cost, he wants to lay theminimum 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 mustfind the minimum amount of fiber needed to connect them all together. Each farmmust connect to some other farm such that a packet can flow from any one farmto any other farm.
The distance between any two farms will not exceed 100,000.

Input

The input includes severalcases. 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, whereeach element shows the distance from on farm to another. Logically, they are N linesof N space-separated integers. Physically, they are limited in length to 80characters, so some lines continue onto others. Of course, the diagonal will be0, since the distance from farm i to itself is not interesting for thisproblem.

Output

For each case, output a singleinteger length that is the sum of the minimum length of fiber required toconnect the entire set of farms.

Sample Input

0 4 9 21

4 0 8 17

9 8 0 16

21 17 16 0

Sample Output

Source

USACO102

解题思路：

给你一张图，求最短路径，这是一到典型的求最短路径模版体，用prim()算法就能过。

代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#define MAX 103
#define VALUE 0xfffff
using namespace std;

int g[MAX][MAX];
int visited[MAX];
int minCost[MAX];
int prim(int n)
{
    int i;
    for(i=0;i<n;i++)
    {
        visited[i]=0;
        minCost[i]=VALUE;
    }
    minCost[0]=0;
    int res=0;
    while(true)
    {
        int t=-1;
        for(i=0;i<n;i++)
        {
            if(visited[i]==0 && (t==-1 || minCost[i]<minCost[t]))
            {
                t=i;
            }
        }
        if(t==-1)
            break;
        visited[t]=1;
        res+=minCost[t];
        for(i=0;i<n;i++)
        {
            if(minCost[i]>g[i][t] && g[i][t]!=0)
            {
                minCost[i]=g[i][t];
            }
        }
    }
    return res;
}

int main()
{
    int i,j;
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(g,0,sizeof(g));
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                scanf("%d",&g[i][j]);
            }
        }
        printf("%d\n",prim(n));
    }
    return 0;
}