夺宝奇兵--组合

作为一名java选手,该oj貌似没有设置java审核时间 结果超时了（对java超级不友好）
不过思路都是一样的 都用贪心解决 因为一个取一个来 1~> n n~> 1 思路如下

import java.util.Scanner;
class Point {
	int X, Y;
}
public class Main {
	public static int Dis(Point Key1, Point Key2) {
		return Math.abs(Key1.X - Key2.X) + Math.abs(Key1.Y - Key2.Y);
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		Point Treasure[][] = new Point[100001][2];
		long Ans = 0;
		for (int i = 1; i <= N; ++i) {
			Treasure[i][0] = new Point();
			Treasure[i][1] = new Point();
			Treasure[i][0].X = sc.nextInt();
			Treasure[i][0].Y = sc.nextInt();
			Treasure[i][1].X = sc.nextInt();
			Treasure[i][1].Y = sc.nextInt();
		}
		for (int i = 1; i < N; ++i) {
			Ans += Math.min(Dis(Treasure[i][0], Treasure[i + 1][0]) + Dis(Treasure[i][1], Treasure[i + 1][1]),
					Dis(Treasure[i][0], Treasure[i + 1][1]) + Dis(Treasure[i][1], Treasure[i + 1][0]));
		}
		Ans += Dis(Treasure[N][0], Treasure[N][1]);
		System.out.println(Ans);
	}
}

#include<bits/stdc++.h>
using namespace std;
struct Point {
    int X, Y;
};
int Dis(Point Key1, Point Key2) {
    return abs(Key1.X - Key2.X) + abs(Key1.Y - Key2.Y);
}
int N, M;
Point Treasure[100010][2];
long long Ans;
int main(int argc, char *argv[]) {
    scanf("%d%d", &N, &M);
    for (int i = 1; i <= N; ++i) {
        scanf("%d%d", &Treasure[i][0].X, &Treasure[i][0].Y);
        scanf("%d%d", &Treasure[i][1].X, &Treasure[i][1].Y);
    }
    for (int i = 1; i < N; ++i) {
        Ans += min(Dis(Treasure[i][0], Treasure[i + 1][0]) + Dis(Treasure[i][1], Treasure[i + 1][1]),
                Dis(Treasure[i][0], Treasure[i + 1][1]) + Dis(Treasure[i][1], Treasure[i + 1][0]));
    }
    Ans += Dis(Treasure[N][0], Treasure[N][1]);
    printf("%lld
", Ans);
    return 0;
}