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package class_2022_10_1_week;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.List;
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// 来自Leetcode周赛
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// 魔物了占领若干据点，这些据点被若干条道路相连接，
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// roads[i] = [x, y] 表示编号 x、y 的两个据点通过一条道路连接。
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// 现在勇者要将按照以下原则将这些据点逐一夺回：
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// 在开始的时候，勇者可以花费资源先夺回一些据点，
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// 初始夺回第 j 个据点所需消耗的资源数量为 cost[j]
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// 接下来，勇者在不消耗资源情况下，
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// 每次可以夺回一个和「已夺回据点」相连接的魔物据点，
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// 并对其进行夺回
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// 为了防止魔物暴动，勇者在每一次夺回据点后（包括花费资源夺回据点后），
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// 需要保证剩余的所有魔物据点之间是相连通的（不经过「已夺回据点」）。
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// 请返回勇者夺回所有据点需要消耗的最少资源数量。
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// 输入保证初始所有据点都是连通的，且不存在重边和自环
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// 测试链接 : https://leetcode.cn/problems/s5kipK/
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public class Code02_CaptureStrongHold {
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public static long minimumCost(int[] cost, int[][] roads) {
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int n = cost.length;
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if (n == 1) {
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return cost[0];
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}
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int m = roads.length;
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DoubleConnectedComponents dc = new DoubleConnectedComponents(n, m, roads);
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long ans = 0;
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// dcc {a,b,c} {c,d,e}
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if (dc.dcc.size() == 1) {
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ans = Integer.MAX_VALUE;
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for (int num : cost) {
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ans = Math.min(ans, num);
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}
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} else { // 不只一个点双连通分量
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ArrayList<Integer> arr = new ArrayList<>();
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for (List<Integer> set : dc.dcc) {
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int cutCnt = 0;
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int curCost = Integer.MAX_VALUE;
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for (int nodes : set) {
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if (dc.cut[nodes]) {
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cutCnt++;
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} else {
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curCost = Math.min(curCost, cost[nodes]);
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}
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}
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if (cutCnt == 1) {
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arr.add(curCost);
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}
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}
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arr.sort((a, b) -> a - b);
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for (int i = 0; i < arr.size() - 1; i++) {
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ans += arr.get(i);
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}
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}
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return ans;
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}
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public static class DoubleConnectedComponents {
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// 链式前向星建图
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public int[] head;
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public int[] next;
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public int[] to;
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public int[] dfn;
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public int[] low;
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public int[] stack;
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public List<List<Integer>> dcc;
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public boolean[] cut;
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public static int edgeCnt;
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public static int dfnCnt;
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public static int top;
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public static int root;
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public DoubleConnectedComponents(int n, int m, int[][] roads) {
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init(n, m);
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createGraph(roads);
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creatDcc(n);
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}
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private void init(int n, int m) {
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head = new int[n];
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Arrays.fill(head, -1);
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next = new int[m << 1];
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to = new int[m << 1];
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dfn = new int[n];
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low = new int[n];
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stack = new int[n];
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dcc = new ArrayList<>();
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cut = new boolean[n];
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edgeCnt = 0;
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dfnCnt = 0;
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top = 0;
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root = 0;
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}
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private void createGraph(int[][] roads) {
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for (int[] edges : roads) {
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add(edges[0], edges[1]);
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add(edges[1], edges[0]);
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}
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}
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private void add(int u, int v) {
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to[edgeCnt] = v;
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next[edgeCnt] = head[u];
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head[u] = edgeCnt++;
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}
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private void creatDcc(int n) {
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for (int i = 0; i < n; i++) {
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// 0 1 2 3 n-1
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if (dfn[i] == 0) {
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root = i;
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tarjan(i);
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}
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}
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}
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private void tarjan(int x) {
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dfn[x] = low[x] = ++dfnCnt;
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stack[top++] = x;
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int flag = 0;
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if (x == root && head[x] == -1) {
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dcc.add(new ArrayList<>());
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dcc.get(dcc.size() - 1).add(x);
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} else {
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// 当前来到的节点是x
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// x {a,b,c}
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for (int i = head[x]; i >= 0; i = next[i]) {
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// y是下级节点
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int y = to[i];
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if (dfn[y] == 0) { // y点没遍历过！
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tarjan(y);
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if (low[y] >= dfn[x]) { // 正在扎口袋
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flag++;
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if (x != root || flag > 1) {
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cut[x] = true;
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}
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List<Integer> curAns = new ArrayList<>();
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// 从栈里一次弹出节点
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// 弹到y停！
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// 弹出的节点都加入集合，x也加入，x不弹出
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for (int z = stack[--top]; z != y; z = stack[--top]) {
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curAns.add(z);
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}
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curAns.add(y);
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curAns.add(x);
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dcc.add(curAns);
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}
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low[x] = Math.min(low[x], low[y]);
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} else { // y点已经遍历过了！
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low[x] = Math.min(low[x], dfn[y]);
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}
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}
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}
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}
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}
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}
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