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package class_2023_03_5_week;
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import java.util.Arrays;
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// 来自小红书、字节跳动
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// 村里面一共有 n 栋房子
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// 我们希望通过建造水井和铺设管道来为所有房子供水。
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// 对于每个房子 i，我们有两种可选的供水方案：
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// 一种是直接在房子内建造水井
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// 成本为 wells[i - 1] （注意 -1 ，因为 索引从0开始 ）
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// 另一种是从另一口井铺设管道引水
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// 数组 pipes 给出了在房子间铺设管道的成本
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// 其中每个 pipes[j] = [house1j, house2j, costj]
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// 代表用管道将 house1j 和 house2j连接在一起的成本。连接是双向的。
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// 请返回 为所有房子都供水的最低总成本 。
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// 这道题很高频，引起注意
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// 本身也不难，转化一下变成最小生成树的问题即可
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// 测试链接 : https://leetcode.cn/problems/optimize-water-distribution-in-a-village/
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public class Code03_OptimizeWaterDistributionInVillage {
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public static int MAXN = 10010;
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public static int[][] edges = new int[MAXN << 1][3];
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public static int esize;
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public static int[] father = new int[MAXN];
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public static int[] size = new int[MAXN];
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public static int[] help = new int[MAXN];
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public static void build(int n) {
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for (int i = 0; i <= n; i++) {
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father[i] = i;
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size[i] = 1;
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}
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}
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public static int find(int i) {
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int s = 0;
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while (i != father[i]) {
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help[s++] = i;
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i = father[i];
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}
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while (s > 0) {
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father[help[--s]] = i;
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}
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return i;
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}
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public static boolean union(int i, int j) {
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int f1 = find(i);
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int f2 = find(j);
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if (f1 != f2) {
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if (size[f1] >= size[f2]) {
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father[f2] = f1;
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size[f1] += size[f2];
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} else {
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father[f1] = f2;
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size[f2] += size[f1];
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}
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return true;
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} else {
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return false;
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}
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}
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public static int minCostToSupplyWater(int n, int[] wells, int[][] pipes) {
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esize = 0;
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for (int i = 0; i < n; i++, esize++) {
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edges[esize][0] = 0;
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edges[esize][1] = i + 1;
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edges[esize][2] = wells[i];
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}
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for (int i = 0; i < pipes.length; i++, esize++) {
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edges[esize][0] = pipes[i][0];
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edges[esize][1] = pipes[i][1];
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edges[esize][2] = pipes[i][2];
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}
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Arrays.sort(edges, 0, esize, (a, b) -> a[2] - b[2]);
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build(n);
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int ans = 0;
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for (int i = 0; i < esize; i++) {
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if (union(edges[i][0], edges[i][1])) {
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ans += edges[i][2];
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}
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}
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return ans;
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}
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}
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