Raucous Rockers~

Raucous 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.

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.

PROGRAM NAME: rockers

INPUT FORMAT

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.

SAMPLE INPUT (file rockers.in)

4 5 24 3 4 2

OUTPUT FORMAT

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

SAMPLE OUTPUT (file rockers.out)

3题意：给你 m 张光盘，每个光盘最多只能刻 t 分钟，现在你有 n 首歌，求最多可以刻多少歌，歌的顺序不能改变。分析：状态压缩或直接DFS方法一：状态压缩

View Code

/*  ID: dizzy_l1  LANG: C++  TASK: rockers*/#include<iostream>#include<cstring>#include<cstdio>#include<cmath>using namespace std;int a[21],p[21],v[21];int n,t,m,cnt,ans,tans;bool judge(){    int i,j;    for(i=0;i<m;i++) v[i]=t;    for(i=0,j=0;i<cnt;i++)    {        if(a[p[i]]>t) return false;        if(a[p[i]]<=v[j]) v[j]-=a[p[i]];        else        {            j++;            v[j]-=a[p[i]];        }        if(j==m) return false;    }    return true;}void work(int x){    int tx,i;    tx=x;    while(tx)    {        i=tx&(-tx);        tx^=i;        i=log2(i*1.0);        p[cnt++]=i;    }    if(cnt<=ans) return ;    if(judge()) ans=cnt;}int main(){    freopen("rockers.in","r",stdin);    freopen("rockers.out","w",stdout);    int nn,i;    while(scanf("%d%d%d",&n,&t,&m)==3)    {        for(i=0;i<n;i++) scanf("%d",&a[i]);        nn=pow(2,n);        ans=0;        for(i=1;i<=nn-1;i++)        {            cnt=0;            work(i);        }        printf("%d\n",ans);    }    return 0;}

方法二：DFS

View Code

 #include<iostream> #include<fstream> #include<cstdio> #include <time.h> #define N 21 using namespace std;  int ans,max,a[N],b[N],n,t,m;  void work(int k,int i,int r) {     if(k==m||i==n)     {         if(ans<max) ans=max;         return ;     }     if(max+r<=ans) return ;//减的巧呀！！！     if(b[k]>=a[i])     {         b[k]-=a[i];         max++;         work(k,i+1,r-1);         max--;         b[k]+=a[i];         work(k,i+1,r-1);     }     else      {         work(k+1,i,r);         work(k,i+1,r-1);         } }  int main() { //    freopen("E.txt","r",stdin); //    freopen("out.txt","w",stdout);     int i;     while(cin>>n>>t>>m)     {         max=ans=0;         for(i=0;i<n;i++)          {             cin>>a[i];                 }         for(i=0;i<m;i++) b[i]=t;         work(0,0,n);         cout<<ans<<endl;     } //    cout<<(double)clock()/CLOCKS_PER_SEC<<endl;     return 0; }

 方法三：dp

d[a][b][c]  表示第 a 个光盘已经使用了 b 分钟最后一首歌是 c。

d[ a ][ b+length[d] ][ d ] =max(d[a][b][c]+1 , d[ a ][ b+length[d] ][ d ])