Sunday, August 21, 2016

USACO rockers – 小溪开大船


USACO rockers – 小溪开大船
你刚刚继承了流行的"破锣摇滚"乐队录制的尚未发表的N(1 <= N <= 20)首歌的版权。你打算从中精选一些歌曲,发行M(1 <= M <= 20)张CD。每一张CD最多可以容纳T(1 <= T <= 20)分钟的音乐,一首歌不能分装在两张CD中。
不巧你是一位古典音乐迷,不懂如何判定这些歌的艺术价值。于是你决定根据以下标准进行选择:
1.歌曲必须按照创作的时间顺序在所有的CD盘上出现。(注:第i张盘的最后一首的创作时间要早于第i+1张盘的第一首)
2.选中的歌曲数目尽可能地多。

格式

PROGRAM NAME: rockers
INPUT FORMAT:
(file rockers.in)
第一行: 三个整数:N, T, M.
第二行: N个整数,分别表示每首歌的长度,按创作时间顺序排列。
OUTPUT FORMAT:
(file rockers.out)
一个整数,表示可以装进M张CD盘的乐曲的最大数目。

SAMPLE INPUT

SAMPLE OUTPUT

此题是很经典的01背包问题变种,需要你对01背包解法做一些变通,这类型的题目有很多,比如有N个抽屉,需要放入M个物品等等,以后遇到这种题目就不怕不怕啦。。。。
int n, t, m;
int songs[20]; // 每首歌曲的长度
// 递归穷搜
// value为目前为止累积刻录的歌曲数量,sindex表示第i首歌,bindex表示第i张光盘,remain表示当前光盘容量
int fill(int value, int sindex, int bindex, int remain)
{
    if(sindex >= n || bindex >= m)
        return value;
    
    // 容量是否充足
    bool enough = remain >= songs[sindex];
    return std::max(fill(value, sindex + 1, bindex, remain),
               enough ?
               fill(value + 1, sindex + 1, bindex, remain - songs[sindex])
               : fill(value, sindex, bindex + 1, t));
}
 
// 非递归穷搜
int search()
{
    int max = 0;
    unsigned int mark = 1;
    unsigned int mark_max = 1 << n;
    // mark的2进制位代表第i张歌曲是否需要拷入cd
    for (; mark < mark_max; mark += 1)
    {
        int value = 0;
        int index = 0;
        int remain = t;
        
        int i = 0;
        for(; i < n; ++i)
        {
            if (mark & (1 << i))
            {
                if(songs[i] > t)
                    break;
                
                if(remain >= songs[i])
                {
                    remain -= songs[i];
                    value += 1;
                }
                else
                {
                    index += 1;
                    if(index >= m)
                        break;
                    remain = t - songs[i];
                    value += 1;
                }
            }
        }
        
        if (i == n && max < value)
            max = value;
    }
    
    return max;
}
 
int jump(int v, int j)
{
    if(v > j) return -1;
    
    return (j % t - v) >= 0 ? (j - v) : (j / t * t - v);
}
 
// 01背包的解法
// 此解法的难度在于如何在容量不足时换碟
// 技巧在于,把背包容量看成m*t大小,划分成m块,每次去判断单个块容量是否够放,如果不够则跳到下一个块
// 这样我们的背包公式可表示为
// f[i][j] = max(f[i-1][j], f[i-1][jump(songs[i], j)] + 1)
// i为第i首歌曲,j为容量,songs[i]表示为歌曲需要消耗的容量,jump函数为跳转到相应的容量位置,
// f[i][j]则表示前i个物品容量小于等于j时的最优解
// 降维打击后可变为1维,公式就不列了
int package01()
{
    
    int values[20*20 + 1] = {0};
    for(int i = 0; i < n; ++i)
    {
        for(int j = m*t; j >= songs[i]; --j)
        {
            int index = jump(songs[i], j);
            values[j] = std::max(values[j], index >= 0 ? values[index]+1 : 0);
 
            if(i == n - 1) break;
        }
    }
    
    return values[m*t];
}
 
int main()
{
    ifstream fin("rockers.in");
    ofstream fout("rockers.out");
    
    fin >> n >> t >> m;
    for(int i = 0; i < n; ++i)
    {
        fin >> songs[i];
    }
    
    // 递归穷搜超时
//    fout << fill(0, 0, 0, t);
    // 非递归穷搜通过
//    fout << search() << endl;
    // 01背包最快最经典
    fout << package01() << endl;
    
    return 0;
}
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