poj1285 Agri-Net Prim最小生成树

最新推荐文章于 2021-01-10 20:16:21 发布

cloudy_happy

最新推荐文章于 2021-01-10 20:16:21 发布

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分类专栏： krustal && Prim 最小生成树

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本文详细解析了如何使用最小生成树算法解决农民John的问题，即通过连接所有农场来建立农业网络，同时最小化光纤铺设成本。文章通过一个具体的样例，介绍了Prim算法的实现过程，展示了如何从邻接矩阵中筛选出最小生成树，从而达到最优的网络布局。

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Agri-Net

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

思路：

先说题意吧，大致就是，给一个点集的邻接矩阵，矩阵中的a[ i ][ j ] = k代表 i 与j 之间相距 k，求最小生成树的最长路，

还是很简单的，跑Prim的时候筛一下就好了

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 110;
int map[maxn][maxn];
int cost[maxn];
bool vis[maxn];

int prim(int n)
{
	memset(cost,inf,sizeof(cost));
	memset(vis,0,sizeof(vis));
	int res = 0,v;
	cost[1] = 0;
	while (1)
	{
		v = -1;
		for (int u = 1;u <= n;u ++)
			if (!vis[u] && (v == -1 || cost[u] < cost[v]))
				v = u;
		vis[v] = 1;
		if (v == -1)
			break;
		res += cost[v];
		for (int u = 1;u <= n;u ++)
		{
			cost[u] = min(cost[u],map[v][u]);
		}
	} 
	return res;
}
int main()
{
	int n;
	while (~scanf("%d",&n))
	{
		for (int i = 1;i <= n;i ++)
			for (int j = 1;j <= n;j ++)
				scanf("%d",&map[i][j]);
		int ans = prim(n);
		printf("%d\n",ans);
	}
	return 0;
}