[POJ]1258 Agri-Net-优快云博客

本文链接：https://blog.youkuaiyun.com/Miner_Sty/article/details/50087725

本文介绍了一个基于Prim算法构建最小生成树的问题，通过实例详细解析了如何寻找连接所有节点的最短路径，适用于网络布局和成本优化等场景。

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[POJ]1258 Agri-Net

问题

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

分析

题目大意是，FJ当上了镇长（怎么又是FJ），现在他想搭建网络，问怎么搭建最节省。
通过构建最小生成树，采用Prim算法。
依旧题中的数据为例。网络初始情况如下图所示。

选取1为起点，则1-2为4，1-3为9，1-4为21，因此选择最小的4，即点2。

选择2后，重构权重图，（1，2）-3为2-3的8，（1，2）-4为2-4的17，因此选择3
选择最小的2之后
选择3后，重构权重图，（1，2，3）-4为3-4的16
选择最小的3之后
最终的最小生成树如下所示：

通过累加最小权重值4+8+16 = 28，因此最小代价为28

源代码

#include <iostream>
#include <vector>
using namespace std;

int Prim(int n, vector<vector<int> >array) {
    int sum = 0, time = 1;
    vector<int> low(n, 0);
    vector<bool> flag(n, true);
    for (int i = 1; i < n; ++i) {
        low[i] = array[0][i];
        flag[i] = false;
    }
    while (time <= n - 1) {
        int min = 99999999;
        int j = 0;
        for (int k = 1; k < n; ++k)
            if (low[k] < min && flag[k] == false) {
                min = low[k];
                j = k;
            }
        sum += min;
        flag[j] = true;
        for (int k = 1; k < n; ++k)
            if (array[j][k] < low[k] && flag[k] == false)
                low[k] = array[j][k];
        ++time;
    }
    return sum;
}

int main() {
    int N;
    while (cin >> N) {
        vector<vector<int> >array(N, vector<int>(N, 0));
        for (int i = 0; i < N; ++i)
            for (int j = 0; j < N; ++j)
                cin >> array[i][j];
        cout << Prim(N, array) << endl;
    }
    return 0;
}