【USACO 3.4】(rockers)Raucous Rockers

------------Prob-------------

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.

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.

--------------Solution--------------

能这样欣赏音乐我六体投地啊。。。

二维DP

f[i][j]表示前i首歌中取j首所花费的CD数和最后一张CD的剩余时间

由问题目标易知应当使f[i][j]尽可能小

显然f[i][j]只有f[i-1][j]和f[i-1][j-1]转移的得到

状态量O(n^2)

转移代价O(1)

时间复杂度O(n^2)

---------------Code------------------

/*
ID:zst_0111
TASK:rockers
LANG:C++
*/
#include<stdio.h>
#include<stdlib.h>
 
int f[22][22][2];
int len[22];
int main()
{
    int i,j,k,m,n;
    int T;
    freopen("rockers.in","r",stdin);
    freopen("rockers.out","w",stdout);
    
    scanf("%d%d%d",&n,&T,&m);
    k=0;
    for(i=1;i<=n;i++)
    {
        scanf("%d",&len[++k]);
        if(len[k]>T)
            k--;
    }
    n=k;
    
    for(i=0;i<=n;i++)
        for(j=0;j<=n;j++)
        {
            f[i][j][0]=214748364;
            f[i][j][1]=T+1;
        }
    for(i=0;i<=n;i++)
    {
        f[i][0][0]=1;
        f[i][0][1]=0;
    }
    for(i=1;i<=n;i++)
        for(j=1;j<=i;j++)
        {
            f[i][j][0]=f[i-1][j][0];
            f[i][j][1]=f[i-1][j][1];
            int fa,fb;
            int ta=f[i-1][j-1][0];
            int tb=f[i-1][j-1][1];
            if(tb+len[i]<=T)
            {
                fa=ta;
                fb=tb+len[i];
                if(fa<f[i][j][0] || fa==f[i][j][0]&&fb<f[i][j][1])
                {
                    f[i][j][0]=fa;
                    f[i][j][1]=fb;
                }
            }
            else
            {
                fa=ta+1;
                fb=len[i];
                if(fa<f[i][j][0] || fa==f[i][j][0]&&fb<f[i][j][1])
                {
                    f[i][j][0]=fa;
                    f[i][j][1]=fb;
                }
            }
        }
    for(i=n;i>=0;i--)
        if(f[n][i][0]<=m)
        {
            printf("%d\n",i);
            return 0;
        }
    return 0;
}