2018阿里编程测验题：最短汉密尔顿回路

Posted 2021-01-02 weiyinfu

N个城市，要求走一条最短的回路

import java.util.Arrays;
import java.util.Scanner;

public class Main {
class Point {
    int x, y;

    Point(int x, int y) {
        this.x = x;
        this.y = y;
    }
}

//最终的答案
int ans = Integer.MAX_VALUE;
//visited[i]表示第i个地点被访问过
boolean visited[];
//存储全部地点
Point[] a;
//起点
Point start = new Point(0, 0);

//求两个点之间的距离
int getDis(Point x, Point y) {
    return Math.abs(x.x - y.x) + Math.abs(x.y - y.y);
}

/**
 * 递归求解
 *
 * @param cur        当前所在位置
 * @param dis        当前已经走过的距离
 * @param visitCount 已经访问的城市个数，用来记录递归终点
 **/
void go(Point cur, int dis, int visitCount) {
    if (visitCount == visited.length) {//如果已经访问完了，更新答案
        ans = Math.min(ans, dis + getDis(start, cur));
    }
    //利用到起点的距离剪枝
    if (dis > ans) {
        return;
    }
    //利用到终点的距离剪枝
    if (dis + getDis(cur, start) > ans) {
        return;
    }
    //对于每个未曾访问的城市进行处理
    for (int i = 0; i < visited.length; i++) {
        if (!visited[i]) {
            visited[i] = true;
            go(a[i], dis + getDis(cur, a[i]), visitCount + 1);
            visited[i] = false;
        }
    }
}

Main() {
    Scanner cin = new Scanner(System.in);
    int n = cin.nextInt();
    a = new Point[n];
    visited = new boolean[n];
    for (int i = 0; i < n; i++) {
        int pos[] = Arrays.stream(cin.next().split(",")).mapToInt(Integer::parseInt).toArray();
        a[i] = new Point(pos[0], pos[1]);
    }
    go(new Point(0, 0), 0, 0);
    System.out.println(ans);
}


public static void main(String[] args) {
    new Main();
}
}