USACO3.4 Raucous Rockers

Raucous RockersRaucous Rockers

You just inherited the rights to N (1 <= N <= 20) previously unreleased songs recorded by the popular group Raucous Rockers. You plan to release a set of M (1 <= M <= 20) compact disks with a selection of these songs. Each disk can hold a maximum of T (1 <= T <= 20) minutes of music, and a song can not overlap from one disk to another.

你刚刚获得了流行乐队Raucous Rockers录制的N（1 <= N <= 20）首未发行歌曲的版权。您计划发布一组M（1 <= M <= 20）张包含这些歌曲的 CD。每张 CD 可以保存最多T（1 <= T <= 20）分钟的音乐，并且一首歌曲不能分段保存。

Since you are a classical music fan and have no way to judge the artistic merits of these songs, you decide on the following criteria for making the selection:

• The songs on the set of disks must appear in the order of the dates that they were written.
• The total number of songs included will be maximized.

由于你是古典音乐迷，无法判断这些歌曲的艺术价值，你决定根据以下标准进行选择：

• CD 上的歌曲必须按照这些歌曲的完成时间依次出现（从较早到较晚）。
• 最大化包含的歌曲总数。

PROGRAM NAME: rockers

INPUT FORMAT 输入格式

Line 1: Three Integers: N, T, and M.

Line 2: N integers that are the lengths of the songs ordered by the date they were written.

第 1 行：三个整数 N，T和M。

第 2 行：N 个整数，表示了每首歌曲的长度（按照完成时间排序）。

SAMPLE INPUT (file rockers.in) 样例输入（文件 rockers.in)

4 5 2
4 3 4 2

OUTPUT FORMAT 输出格式

A single line with an integer that is the number of songs that will fit on M disks.

SAMPLE OUTPUT (file rockers.out) 样例输出（文件 rockers.out）

3

题解

类似背包；上代码：

/*
ID: hanghan1
LANG: C++11
PROB: rockers
*/
#define _TASK_ "rockers"
#include <bits/stdc++.h>
using namespace std;
#define N 105
int dp[N][N], len[N], f[N];

int main() {
    freopen(_TASK_ ".in", "r", stdin);
    freopen(_TASK_ ".out", "w", stdout);
    int n, t, m; scanf("%d%d%d", &n,&t,&m);
    for (int i=1; i<=n; i++) scanf("%d", &len[i]);
    for (int i=1; i<=n; i++) {
        for (int j=m; j>=1; j--) {
            dp[j][0] = f[j-1];
            for (int k=t; k>=len[i]; k--) if (dp[j][k] < dp[j][k-len[i]]+1) {
                dp[j][k] = dp[j][k-len[i]]+1;
                f[j] = max(f[j], dp[j][k]);
            }
        }
    }
    printf("%d\n", f[m]);
    return 0;
}