*[topcoder]GUMIAndSongsDiv1

http://community.topcoder.com/stat?c=problem_statement&pm=12706&rd=15700

这题有意思。首先要观察到，如果选定一些歌曲，最优做法就是，按照tone排序，那么这时浪费的间隔最少，是(max_tone-min_tone)。那么首先对歌曲按照tone排序，这时由于取得顺序就是从左往右，可以用DP。（比如：http://community.topcoder.com/stat?c=problem_solution&cr=23061369&rd=15700&pm=12706）但又其实对于给定的歌曲集，用贪心按照duration从小到大取就行，那么用n*n来遍历歌曲集的选取，然后用贪心选至超过T就行了。

import java.util.*;

class Song {
    int duration;
    int tone;
    public Song(int duration, int tone) {
        this.duration = duration;
        this.tone = tone;
    }
}

public class GUMIAndSongsDiv1 {
    public int maxSongs(int[] duration, int[] tone, int T) {
        int len = duration.length;
        Song[] songs = new Song[len];
        for (int i = 0; i < len; i++) {
            songs[i] = new Song(duration[i], tone[i]);
        }
        Arrays.sort(songs, new Comparator<Song>() {
            public int compare(Song a, Song b) {
                return a.tone - b.tone;
            }
        });
        int res = 0;
        for (int first = 0; first < len; first++) {
            for (int last = first; last < len; last++) {
                Song[] tmp = new Song[last - first + 1];
                System.arraycopy(songs, first, tmp, 0, tmp.length);
                int timeLeft = T - (songs[last].tone - songs[first].tone);
                Arrays.sort(tmp, new Comparator<Song>() {
                    public int compare(Song a, Song b) {
                        return a.duration - b.duration;
                    }
                });
                int cnt = 0;
                for (int i = 0; i < tmp.length; i++) {
                    if (tmp[i].duration <= timeLeft) {
                        cnt++;
                        timeLeft -= tmp[i].duration;
                    }
                }
                res = Math.max(res, cnt);
            }
        }
        return res;
    }
}