工作分配

题目：
  有最多15件事，分给两个人做。一个人不能同时做两件事，即分给同一个人的事情在时间上不能重叠。求两个人最多可以做多长时间的事情。

要求：
  每件事情有开始时间和结束时间，时间由时、分两部分构成，输出以分钟为单位。

import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;

public class Working {

static int[][] works;static int[] seq;static int n;static int max;public static void main(String[] args) throws FileNotFoundException {    // TODO Auto-generated method stub    @SuppressWarnings("resource")    Scanner sc = new Scanner(System.in);    sc = new Scanner(new File("files/working"));    int T = sc.nextInt();    for (int t = 0; t < T; t++) {        n = sc.nextInt();        works = new int[n][3];        seq=new int[n];        int hour, sec;        for (int i = 0; i < n; i++) {            hour = sc.nextInt();            sec = sc.nextInt();            works[i][0] = hour * 60 + sec;            hour = sc.nextInt();            sec = sc.nextInt();            works[i][1] = hour * 60 + sec;            works[i][2] = works[i][1] - works[i][0];        }        Qsort(0, n - 1, works);        max = 0;        dfs(0, 0, 0, 0);        System.out.println(max);    }}private static void dfs(int step, int time, int last1, int last2) {    // TODO Auto-generated method stub    if (step == n) {        if (time > max) {            max = time;        }        return;    }    if(Res()<=max-time)        return;    // 嫁给第一个人    if (works[step][0] >= last1) {        seq[step]=1;        dfs(step + 1, time + works[step][2], works[step][1], last2);    }    // 嫁给第二个人    if (works[step][0] >= last2) {        seq[step]=2;        dfs(step + 1, time + works[step][2], last1, works[step][1]);    }    // 孤独终生    seq[step]=3;    dfs(step + 1, time, last1, last2);    seq[step]=0;}private static int Res() {    // TODO Auto-generated method stub    int sum=0;    for(int i=0;i<n;i++){        if(seq[i]==0)            sum+=works[i][2];    }    return sum;}private static void Qsort(int l, int r, int[][] w) {    // TODO Auto-generated method stub    if (l < r) {        int[] key = new int[3];        CopyOf(key, w[r]);        int i = l;        int j = r;        while (i < j) {            while (i < j && w[i][0] < key[0])                i++;            CopyOf(w[j], w[i]);            while (i < j && w[j][0] >= key[0])                j--;            CopyOf(w[i], w[j]);        }        CopyOf(w[j], key);        Qsort(l, j - 1, w);        Qsort(j + 1, r, w);    }}private static void CopyOf(int[] s1, int[] s2) {    // TODO Auto-generated method stub    for (int i = 0; i < s1.length; i++) {        s1[i] = s2[i];    }}

}

sample input:
5
5
8 0 14 0
9 0 12 0
13 0 18 0
10 0 17 0
10 30 13 30
6
8 0 18 10
13 10 15 20
10 50 13 30
13 20 18 10
12 0 12 50
12 0 15 30
8
10 40 12 20
9 10 15 10
11 0 16 30
11 50 15 50
8 30 13 50
15 0 18 10
9 30 10 10
13 10 18 30
10
10 50 13 50
9 40 12 0
8 20 15 10
8 30 10 20
11 40 13 20
8 20 13 50
14 20 18 10
15 0 16 40
8 0 16 50
10 0 11 10
14
10 40 15 50
9 40 10 0
12 40 16 10
12 40 13 0
12 20 17 40
11 40 17 30
9 0 12 0
16 10 17 30
17 0 18 40
8 40 12 40
8 20 14 30
12 40 18 40
15 0 16 20
9 40 10 10
sample output:
900–[0, 1, 2, 0, 1]
950–[1, 0, 0, 2, 0, 2]
970–[1, 0, 2, 2, 0, 0, 2, 1]
1090–[1, 2, 0, 0, 0, 0, 0, 0, 2, 0]
1150–[1, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 1]