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English Version

题目描述

想象一下你是个城市基建规划者,地图上有 n 座城市,它们按以 1 到 n 的次序编号。

给你整数 n 和一个数组 conections,其中 connections[i] = [xi, yi, costi] 表示将城市 xi 和城市 yi 连接所要的costi连接是双向的)。

返回连接所有城市的最低成本,每对城市之间至少有一条路径。如果无法连接所有 n 个城市,返回 -1

最小成本 应该是所用全部连接成本的总和。

 

示例 1:

输入:n = 3, conections = [[1,2,5],[1,3,6],[2,3,1]]
输出:6
解释:选出任意 2 条边都可以连接所有城市,我们从中选取成本最小的 2 条。

示例 2:

输入:n = 4, conections = [[1,2,3],[3,4,4]]
输出:-1
解释:即使连通所有的边,也无法连接所有城市。

 

提示:

  • 1 <= n <= 104
  • 1 <= connections.length <= 104
  • connections[i].length == 3
  • 1 <= xi, yi <= n
  • xi != yi
  • 0 <= costi <= 105

解法

最小生成树问题。设 n 表示点数,m 表示边数。

方法一:Kruskal 算法

时间复杂度 O(mlogm)。

Python3

class Solution:
    def minimumCost(self, n: int, connections: List[List[int]]) -> int:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        connections.sort(key=lambda x: x[2])
        p = list(range(n))
        ans = 0
        for x, y, cost in connections:
            x, y = x - 1, y - 1
            if find(x) == find(y):
                continue
            p[find(x)] = find(y)
            ans += cost
            n -= 1
            if n == 1:
                return ans
        return -1

Java

class Solution {
    private int[] p;

    public int minimumCost(int n, int[][] connections) {
        Arrays.sort(connections, Comparator.comparingInt(a -> a[2]));
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        int ans = 0;
        for (int[] e : connections) {
            int x = e[0] - 1, y = e[1] - 1, cost = e[2];
            if (find(x) == find(y)) {
                continue;
            }
            p[find(x)] = find(y);
            ans += cost;
            if (--n == 1) {
                return ans;
            }
        }
        return -1;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

C++

class Solution {
public:
    vector<int> p;

    int minimumCost(int n, vector<vector<int>>& connections) {
        p.resize(n);
        for (int i = 0; i < n; ++i) p[i] = i;
        sort(connections.begin(), connections.end(), [](auto& a, auto& b) { return a[2] < b[2]; });
        int ans = 0;
        for (auto& e : connections) {
            int x = e[0] - 1, y = e[1] - 1, cost = e[2];
            if (find(x) == find(y)) continue;
            p[find(x)] = find(y);
            ans += cost;
            if (--n == 1) return ans;
        }
        return -1;
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }
};

Go

func minimumCost(n int, connections [][]int) int {
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	sort.Slice(connections, func(i, j int) bool {
		return connections[i][2] < connections[j][2]
	})
	var find func(x int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	ans := 0
	for _, e := range connections {
		x, y, cost := e[0]-1, e[1]-1, e[2]
		if find(x) == find(y) {
			continue
		}
		p[find(x)] = find(y)
		ans += cost
		n--
		if n == 1 {
			return ans
		}
	}
	return -1
}

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