acm训练2019，1，28 POJ - 2421 建路最小生成树

最新推荐文章于 2022-02-10 22:17:49 发布

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最新推荐文章于 2022-02-10 22:17:49 发布

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分类专栏：最小生成树

本文链接：https://blog.youkuaiyun.com/weixin_44012730/article/details/86684514

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本文深入探讨了最小生成树算法的实现，特别是在已有部分道路连接的村庄网络中，如何通过算法构建剩余道路以确保所有村庄连通且总道路长度最短。文章提供了详细的算法步骤，包括初始化村庄间的距离矩阵，应用并查集数据结构进行节点连接，以及如何处理已存在的道路条件。

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There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
Input
The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
Output
You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.
Sample Input
3
0 990 692
990 0 179
692 179 0
1
1 2
Sample Output
179

基础的最小生成树，多了一个已建成的路条件，先将已建成的加进去，再使用最小生成树

#include <iostream>
#include <algorithm>
using namespace std;
int parent[100005];
struct node {
	int x, y;
	int dis;
};
node qu[10000];

int find_root(int x) {
	int x_root = x;
	while (parent[x_root] != -1) {
		x_root = parent[x_root];
	}
	return x_root;
}

int union_v(int x, int y) {
	int x_root = find_root(x);
	int y_root = find_root(y);
	if (x_root == y_root)return 0;

	else {
		parent[x_root] = y_root;
		return 1;
	}
}

int min(node a, node b) {
	return a.dis < b.dis;
}

int t = 0;

int  main() {
	int n;
	cin >> n;
	memset(parent, -1, sizeof(parent));
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			int temp;
			if (j == i)cin>>temp;
			else {
				qu[t].x = i, qu[t].y = j;
				cin >> qu[t].dis;
				t++;
			}
		}
	}
	sort(qu, qu + t, min);

	int m;
	cin >> m;
	while (m--) {
		int a1, b1;
		cin >> a1 >> b1;
		union_v(a1, b1);
	}

	int ans = 0;
	for (int i = 0; i < t; i++) {
		if (union_v(qu[i].x, qu[i].y)) {
			ans += qu[i].dis;
		}
	}

	cout << ans << endl;
	return 0;
}