(最小生成树-Prim)PKU-1258 Agri-Net-优快云博客

本文介绍了一种使用最小生成树算法解决农业区域互联网连接问题的方法。通过Prim算法实现光纤网络布局，确保所有农场都能以最低成本接入互联网。

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原题链接 http://poj.org/problem?id=1258

Agri-Net

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

Sample Output

<span style="font-size:24px;color:#333399;">Source Code</span>

/*
最小生成树 Prim
*/

#include <iostream>
using namespace std;

#define Max 102
#define INF 1000000;
int map[Max][Max];

int prim(int points) {

	int i, j, k, MaxDis, result;
	bool searched[Max];
	int dis[Max];

	//初始化，以0为起点
	result = 0;
	memset(searched, 0, sizeof(searched));
	for (i = 1; i < points; i++) {
		dis[i] = map[0][i];
	}
	
	searched[0] = 1;
	k = 0;
	for (i = 1; i < points; i++) {
		MaxDis = INF;
		//每次找出最近的点
		for (j = 0; j < points; j++) {
			if (!searched[j] && MaxDis > dis[j]) {
				MaxDis = dis[j];
				k = j;
			} 
		}

		//更新当前点集到其余未加入的点的最短距离
	    searched[k] = 1;
		result += MaxDis;
		for (j = 0; j < points; j++) {
			if (!searched[j] && map[k][j] < dis[j]) {
				dis[j] = map[k][j];
			}
		}
	}

	return result;
}

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

}