【poj1258】Agri-Net——Kruskal

最新推荐文章于 2023-11-30 13:38:17 发布

CFhM_R

最新推荐文章于 2023-11-30 13:38:17 发布

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分类专栏：题解心得最小生成树文章标签： kruskal poj

本文链接：https://blog.youkuaiyun.com/CFhM_R/article/details/50933819

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最小生成树

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本文介绍了一道关于寻找最小生成树的经典算法问题，并通过Kruskal算法结合优先队列进行优化解决，旨在帮助读者理解如何高效地构建连接所有节点的最小代价路径。

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题目：

Agri-Net

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48503		Accepted: 20093

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

Source

USACO 102

描述：给定邻接矩阵，求最小生成树

题解：kruskal裸题，不过这个kruskal是经过堆优化的

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
const int maxn = 105;
int map[maxn][maxn];
struct ee
{
	int u;
	int v;
	int w;
};

int fa[maxn], N;

struct cmp
{
	bool operator()(ee a, ee b)
	{
		return a.w > b.w;
	}
};

priority_queue<ee, vector<ee>, cmp> que;
void init()
{
	for (int i = 1; i <= N; i++)
	{
		fa[i] = i;
	}
	while (!que.empty())
		que.pop();
}
int find(int x)
{
	if (fa[x] == x)
		return x;
	else
		return fa[x] = find(fa[x]);
}

bool unite(int x, int y)
{
	int fx = find(x);
	int fy = find(y);
	if (fx != fy)
	{
		fa[fy] = fx;
		return true;
	}
	return false;
}

int main()
{
	//freopen("input.txt", "r", stdin);
	while (scanf("%d", &N) != EOF)
	{
		for (int i = 1; i <= N; i++)
		{
			for (int j = 1; j <= N; j++)
			{
				scanf("%d", &map[i][j]);
			}
		}
		ee edge1;
		int mnt = 0;
		init();
		for (int i = 1; i <= N; i++)
		{
			for (int j = i; j <= N; j++)
			{
				if (map[i][j] != 0)
				{
					edge1.u = i;
					edge1.v = j;
					edge1.w = map[i][j];
					que.push(edge1);
					mnt++;
				}
			}
		}
		int sum = 0;
		int cnt = 0;
		while (cnt < N - 1)
		{
			ee temp;
			temp = que.top();
			que.pop();
			if (unite(temp.u, temp.v))
			{
				sum += temp.w;
				cnt++;
			}
		}
		printf("%d\n", sum);
	}
	return 0;
}