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package class_2022_12_4_week;
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// Floyd算法应用
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// 存在一个由 n 个节点组成的无向连通图，图中的节点按从 0 到 n - 1 编号
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// 给你一个数组 graph 表示这个图
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// 其中，graph[i] 是一个列表，由所有与节点 i 直接相连的节点组成
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// 返回能够访问所有节点的最短路径的长度
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// 你可以在任一节点开始和停止，也可以多次重访节点，并且可以重用边
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// 测试链接 : https://leetcode.cn/problems/shortest-path-visiting-all-nodes/
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public class Code03_ShortestPathVisitingAllNodes {
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public static int shortestPathLength(int[][] graph) {
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int n = graph.length;
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int[][] distance = floyd(n, graph);
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int ans = Integer.MAX_VALUE;
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int[][] dp = new int[1 << n][n];
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for (int i = 0; i < (1 << n); i++) {
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for (int j = 0; j < n; j++) {
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dp[i][j] = -1;
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}
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}
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for (int i = 0; i < n; i++) {
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ans = Math.min(ans, process(1 << i, i, n, distance, dp));
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}
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return ans;
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}
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public static int[][] floyd(int n, int[][] graph) {
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int[][] distance = new int[n][n];
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// 初始化认为都没路
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < n; j++) {
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distance[i][j] = Integer.MAX_VALUE;
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}
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}
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// 自己到自己的距离为0
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for (int i = 0; i < n; i++) {
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distance[i][i] = 0;
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}
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// 支持任意有向图，把直接边先填入
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for (int cur = 0; cur < n; cur++) {
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for (int next : graph[cur]) {
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distance[cur][next] = 1;
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distance[next][cur] = 1;
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}
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}
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// O(N^3)的过程
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// 枚举每个跳板
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// 注意! 跳板要最先枚举，然后是from和to
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for (int jump = 0; jump < n; jump++) {
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for (int from = 0; from < n; from++) {
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for (int to = 0; to < n; to++) {
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if (distance[from][jump] != Integer.MAX_VALUE && distance[jump][to] != Integer.MAX_VALUE
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&& distance[from][to] > distance[from][jump] + distance[jump][to]) {
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distance[from][to] = distance[from][jump] + distance[jump][to];
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}
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}
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}
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}
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return distance;
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}
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// status : 所有已经走过点的状态
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// 0 1 2 3 4 5
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// int
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// 5 4 3 2 1 0
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// 0 0 1 1 0 1
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// cur : 当前所在哪个城市！
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// n : 一共有几座城
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// 返回值 : 从此时开始，逛掉所有的城市，至少还要走的路，返回！
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public static int process(int status, int cur, int n, int[][] distance, int[][] dp) {
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// 5 4 3 2 1 0
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// 1 1 1 1 1 1
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// 1 << 6 - 1
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if (status == (1 << n) - 1) {
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return 0;
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}
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if (dp[status][cur] != -1) {
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return dp[status][cur];
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}
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int ans = Integer.MAX_VALUE;
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// status:
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// 5 4 3 2 1 0
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// 0 0 1 0 1 1
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// cur : 0
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// next : 2 4 5
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for (int next = 0; next < n; next++) {
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if ((status & (1 << next)) == 0 && distance[cur][next] != Integer.MAX_VALUE) {
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int nextAns = process(status | (1 << next), next, n, distance, dp);
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if (nextAns != Integer.MAX_VALUE) {
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ans = Math.min(ans, distance[cur][next] + nextAns);
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}
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}
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}
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dp[status][cur] = ans;
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return ans;
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}
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}
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