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困难
2069
第 7 场双周赛 Q4
并查集
最小生成树
堆(优先队列)

English Version

题目描述

村里面一共有 n 栋房子。我们希望通过建造水井和铺设管道来为所有房子供水。

对于每个房子 i,我们有两种可选的供水方案:一种是直接在房子内建造水井,成本为 wells[i - 1] (注意 -1 ,因为 索引从0开始 );另一种是从另一口井铺设管道引水,数组 pipes 给出了在房子间铺设管道的成本,其中每个 pipes[j] = [house1j, house2j, costj] 代表用管道将 house1j 和 house2j连接在一起的成本。连接是双向的。

请返回 为所有房子都供水的最低总成本

 

示例 1:

输入:n = 3, wells = [1,2,2], pipes = [[1,2,1],[2,3,1]]
输出:3
解释: 
上图展示了铺设管道连接房屋的成本。
最好的策略是在第一个房子里建造水井(成本为 1),然后将其他房子铺设管道连起来(成本为 2),所以总成本为 3。

示例 2:

输入:n = 2, wells = [1,1], pipes = [[1,2,1]]
输出:2
解释:我们可以用以下三种方法中的一种来提供低成本的水:
选项1:
在1号房子里面建一口井,成本为1
在房子2内建造井,成本为1
总成本是2。
选项2:
在1号房子里面建一口井,成本为1
-花费1连接房子2和房子1。
总成本是2。
选项3:
在房子2内建造井,成本为1
-花费1连接房子1和房子2。
总成本是2。
注意,我们可以用cost 1或cost 2连接房子1和房子2,但我们总是选择最便宜的选项。

 

提示:

  • 2 <= n <= 104
  • wells.length == n
  • 0 <= wells[i] <= 105
  • 1 <= pipes.length <= 104
  • pipes[j].length == 3
  • 1 <= house1j, house2j <= n
  • 0 <= costj <= 105
  • house1j != house2j

解法

方法一:Kruskal 算法(最小生成树)

我们假设有一个水井编号为 $0$,那么我们可以将每个房子与水井 $0$ 之间的连通性看作是一条边,每条边的权值为该房子建造水井的成本。同时,我们将每个房子之间的连通性也看作是一条边,每条边的权值为铺设管道的成本。这样一来,我们就可以将本题转化成求一张无向图的最小生成树的问题。

我们可以使用 Kruskal 算法求出无向图的最小生成树。我们先把水井 $0$ 与房子之间的一条边加入 $pipes$ 数组中,然后将 $pipes$ 数组按照边权值从小到大排序。随后,我们遍历每一条边,如果这条边连接了不同的连通分量,我们就选用这条边,并将对应连通分量合并。如果当前的连通分量恰好为 $1$,那么我们就找到了最小生成树,此时的答案即为当前边权值,我们将其返回即可。

时间复杂度 $O((m + n) \times \log (m + n))$,空间复杂度 $O(m + n)$。其中 $m$$n$ 分别是 $pipes$ 数组和 $wells$ 数组的长度。

Python3

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

        for i, w in enumerate(wells, 1):
            pipes.append([0, i, w])
        pipes.sort(key=lambda x: x[2])
        p = list(range(n + 1))
        ans = 0
        for a, b, c in pipes:
            pa, pb = find(a), find(b)
            if pa != pb:
                p[pa] = pb
                n -= 1
                ans += c
                if n == 0:
                    return ans

Java

class Solution {
    private int[] p;

    public int minCostToSupplyWater(int n, int[] wells, int[][] pipes) {
        int[][] nums = Arrays.copyOf(pipes, pipes.length + n);
        for (int i = 0; i < n; i++) {
            nums[pipes.length + i] = new int[] {0, i + 1, wells[i]};
        }
        Arrays.sort(nums, (a, b) -> a[2] - b[2]);
        p = new int[n + 1];
        for (int i = 0; i <= n; i++) {
            p[i] = i;
        }
        int ans = 0;
        for (var x : nums) {
            int a = x[0], b = x[1], c = x[2];
            int pa = find(a), pb = find(b);
            if (pa != pb) {
                p[pa] = pb;
                ans += c;
                if (--n == 0) {
                    return ans;
                }
            }
        }
        return ans;
    }

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

C++

class Solution {
public:
    int minCostToSupplyWater(int n, vector<int>& wells, vector<vector<int>>& pipes) {
        for (int i = 0; i < n; ++i) {
            pipes.push_back({0, i + 1, wells[i]});
        }
        sort(pipes.begin(), pipes.end(), [](const vector<int>& a, const vector<int>& b) {
            return a[2] < b[2];
        });
        int p[n + 1];
        iota(p, p + n + 1, 0);
        function<int(int)> find = [&](int x) {
            if (p[x] != x) {
                p[x] = find(p[x]);
            }
            return p[x];
        };
        int ans = 0;
        for (const auto& x : pipes) {
            int pa = find(x[0]), pb = find(x[1]);
            if (pa == pb) {
                continue;
            }
            p[pa] = pb;
            ans += x[2];
            if (--n == 0) {
                break;
            }
        }
        return ans;
    }
};

Go

func minCostToSupplyWater(n int, wells []int, pipes [][]int) (ans int) {
	for i, w := range wells {
		pipes = append(pipes, []int{0, i + 1, w})
	}
	sort.Slice(pipes, func(i, j int) bool { return pipes[i][2] < pipes[j][2] })
	p := make([]int, n+1)
	for i := range p {
		p[i] = i
	}
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}

	for _, x := range pipes {
		pa, pb := find(x[0]), find(x[1])
		if pa == pb {
			continue
		}
		p[pa] = pb
		ans += x[2]
		n--
		if n == 0 {
			break
		}
	}
	return
}

TypeScript

function minCostToSupplyWater(n: number, wells: number[], pipes: number[][]): number {
    for (let i = 0; i < n; ++i) {
        pipes.push([0, i + 1, wells[i]]);
    }
    pipes.sort((a, b) => a[2] - b[2]);
    const p = Array(n + 1)
        .fill(0)
        .map((_, i) => i);
    const find = (x: number): number => {
        if (p[x] !== x) {
            p[x] = find(p[x]);
        }
        return p[x];
    };
    let ans = 0;
    for (const [a, b, c] of pipes) {
        const pa = find(a);
        const pb = find(b);
        if (pa === pb) {
            continue;
        }
        p[pa] = pb;
        ans += c;
        if (--n === 0) {
            break;
        }
    }
    return ans;
}

Rust

struct UnionFind {
    p: Vec<usize>,
    size: Vec<usize>,
}

impl UnionFind {
    fn new(n: usize) -> Self {
        let p: Vec<usize> = (0..n).collect();
        let size = vec![1; n];
        UnionFind { p, size }
    }

    fn find(&mut self, x: usize) -> usize {
        if self.p[x] != x {
            self.p[x] = self.find(self.p[x]);
        }
        self.p[x]
    }

    fn union(&mut self, a: usize, b: usize) -> bool {
        let pa = self.find(a);
        let pb = self.find(b);
        if pa == pb {
            false
        } else if self.size[pa] > self.size[pb] {
            self.p[pb] = pa;
            self.size[pa] += self.size[pb];
            true
        } else {
            self.p[pa] = pb;
            self.size[pb] += self.size[pa];
            true
        }
    }
}

impl Solution {
    pub fn min_cost_to_supply_water(n: i32, wells: Vec<i32>, pipes: Vec<Vec<i32>>) -> i32 {
        let n = n as usize;
        let mut pipes = pipes;
        for i in 0..n {
            pipes.push(vec![0, (i + 1) as i32, wells[i]]);
        }
        pipes.sort_by(|a, b| a[2].cmp(&b[2]));
        let mut uf = UnionFind::new(n + 1);
        let mut ans = 0;
        for pipe in pipes {
            let a = pipe[0] as usize;
            let b = pipe[1] as usize;
            let c = pipe[2];
            if uf.union(a, b) {
                ans += c;
                if n == 0 {
                    break;
                }
            }
        }
        ans
    }
}

方法二

Python3

class UnionFind:
    __slots__ = ("p", "size")

    def __init__(self, n):
        self.p = list(range(n))
        self.size = [1] * n

    def find(self, x: int) -> int:
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]

    def union(self, a: int, b: int) -> bool:
        pa, pb = self.find(a), self.find(b)
        if pa == pb:
            return False
        if self.size[pa] > self.size[pb]:
            self.p[pb] = pa
            self.size[pa] += self.size[pb]
        else:
            self.p[pa] = pb
            self.size[pb] += self.size[pa]
        return True


class Solution:
    def minCostToSupplyWater(
        self, n: int, wells: List[int], pipes: List[List[int]]
    ) -> int:
        for i, w in enumerate(wells, 1):
            pipes.append([0, i, w])
        pipes.sort(key=lambda x: x[2])
        uf = UnionFind(n + 1)
        ans = 0
        for a, b, c in pipes:
            if uf.union(a, b):
                ans += c
                n -= 1
                if n == 0:
                    return ans

Java

class UnionFind {
    private int[] p;
    private int[] size;

    public UnionFind(int n) {
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
    }

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

    public boolean union(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            return false;
        }
        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        return true;
    }
}

class Solution {
    public int minCostToSupplyWater(int n, int[] wells, int[][] pipes) {
        int[][] nums = Arrays.copyOf(pipes, pipes.length + n);
        for (int i = 0; i < n; i++) {
            nums[pipes.length + i] = new int[] {0, i + 1, wells[i]};
        }
        Arrays.sort(nums, (a, b) -> a[2] - b[2]);
        UnionFind uf = new UnionFind(n + 1);
        int ans = 0;
        for (var x : nums) {
            int a = x[0], b = x[1], c = x[2];
            if (uf.union(a, b)) {
                ans += c;
                if (--n == 0) {
                    break;
                }
            }
        }
        return ans;
    }
}

C++

class UnionFind {
public:
    UnionFind(int n) {
        p = vector<int>(n);
        size = vector<int>(n, 1);
        iota(p.begin(), p.end(), 0);
    }

    bool unite(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            return false;
        }
        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        return true;
    }

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

private:
    vector<int> p, size;
};

class Solution {
public:
    int minCostToSupplyWater(int n, vector<int>& wells, vector<vector<int>>& pipes) {
        for (int i = 0; i < n; ++i) {
            pipes.push_back({0, i + 1, wells[i]});
        }
        sort(pipes.begin(), pipes.end(), [](const vector<int>& a, const vector<int>& b) {
            return a[2] < b[2];
        });
        UnionFind uf(n + 1);
        int ans = 0;
        for (const auto& x : pipes) {
            if (uf.unite(x[0], x[1])) {
                ans += x[2];
                if (--n == 0) {
                    break;
                }
            }
        }
        return ans;
    }
};

Go

type unionFind struct {
	p, size []int
}

func newUnionFind(n int) *unionFind {
	p := make([]int, n)
	size := make([]int, n)
	for i := range p {
		p[i] = i
		size[i] = 1
	}
	return &unionFind{p, size}
}

func (uf *unionFind) find(x int) int {
	if uf.p[x] != x {
		uf.p[x] = uf.find(uf.p[x])
	}
	return uf.p[x]
}

func (uf *unionFind) union(a, b int) bool {
	pa, pb := uf.find(a), uf.find(b)
	if pa == pb {
		return false
	}
	if uf.size[pa] > uf.size[pb] {
		uf.p[pb] = pa
		uf.size[pa] += uf.size[pb]
	} else {
		uf.p[pa] = pb
		uf.size[pb] += uf.size[pa]
	}
	return true
}

func minCostToSupplyWater(n int, wells []int, pipes [][]int) (ans int) {
	for i, w := range wells {
		pipes = append(pipes, []int{0, i + 1, w})
	}
	sort.Slice(pipes, func(i, j int) bool { return pipes[i][2] < pipes[j][2] })
	uf := newUnionFind(n + 1)
	for _, x := range pipes {
		if uf.union(x[0], x[1]) {
			ans += x[2]
			n--
			if n == 0 {
				break
			}
		}
	}
	return
}

TypeScript

class UnionFind {
    private p: number[];
    private size: number[];

    constructor(n: number) {
        this.p = Array(n)
            .fill(0)
            .map((_, i) => i);
        this.size = Array(n).fill(1);
    }

    find(x: number): number {
        if (this.p[x] !== x) {
            this.p[x] = this.find(this.p[x]);
        }
        return this.p[x];
    }

    union(a: number, b: number): boolean {
        const pa = this.find(a);
        const pb = this.find(b);
        if (pa === pb) {
            return false;
        }
        if (this.size[pa] > this.size[pb]) {
            this.p[pb] = pa;
            this.size[pa] += this.size[pb];
        } else {
            this.p[pa] = pb;
            this.size[pb] += this.size[pa];
        }
        return true;
    }
}

function minCostToSupplyWater(n: number, wells: number[], pipes: number[][]): number {
    for (let i = 0; i < n; ++i) {
        pipes.push([0, i + 1, wells[i]]);
    }
    pipes.sort((a, b) => a[2] - b[2]);
    const uf = new UnionFind(n + 1);
    let ans = 0;
    for (const [a, b, c] of pipes) {
        if (uf.union(a, b)) {
            ans += c;
            if (--n === 0) {
                break;
            }
        }
    }
    return ans;
}