You have a long drive by car ahead. You have a tape recorder, but unfortunately your best music is on

CDs. You need to have it on tapes so the problem to solve is: you have a tape N minutes long. How

to choose tracks from CD to get most out of tape space and have as short unused space as possible.

Assumptions:

• number of tracks on the CD does not exceed

• no track is longer than N minutes

• tracks do not repeat

• length of each track is expressed as an integer number

• N is also integer

Program should find the set of tracks which fills the tape best and print it in the same sequence as

the tracks are stored on the CD

Input

Any number of lines. Each one contains value N, (after space) number of tracks and durations of the

tracks. For example from first line in sample data: N = , number of tracks=, first track lasts for

minute, second one  minutes, next one  minutes

Output

Set of tracks (and durations) which are the correct solutions and string ‘sum:’ and sum of duration

times.

Sample Input

Sample Output

  sum:

  sum:

   sum:

        sum:

      sum:

【代码】：

【一维 + vis标记数组 + 倒序输出】：用一个vis[i][j]记录容量为j的背包里面有没有用到过i物品，物品是倒着放的。

#include<cstdio>

#include<string>

#include<cstdlib>

#include<cmath>

#include<iostream>

#include<cstring>

#include<set>

#include<queue>

#include<algorithm>

#include<vector>

#include<map>

#include<cctype>

#include<stack>

#include<sstream>

#include<list>

#include<assert.h>

#include<bitset>

#include<numeric>

#define debug() puts("++++")

#define gcd(a,b) __gcd(a,b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a,b,sizeof(a))

#define sz size()

#define be begin()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

#define all 1,n,1

#define rep(i,n,x) for(int i=(x); i<(n); i++)

#define in freopen("in.in","r",stdin)

#define out freopen("out.out","w",stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int,int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e18;

const int maxn =  ;

const int maxm = 1e5 + ;

const double PI = acos(-1.0);

const double eps = 1e-;

const int dx[] = {-,,,,,,-,-};

const int dy[] = {,,,-,,-,,-};

const int mon[] = {, , , , , , , , , , , , };

const int monn[] = {, , , , , , , , , , , , };

int n,m;

int v[maxm], w[maxm],dp[maxm];

int t;

bool vis[][maxm];

int main()

{

    while(cin>>m>>n){

        memset(vis,,sizeof(vis));

        memset(dp,,sizeof(dp));

        for(int i=;i<=n;i++)

            cin>>w[i];

        for(int i=n;i>=;i--){ //输出倒序

            for(int j=m;j>=w[i];j--){

                 //dp[j]=max(dp[j],dp[j-w[i]]+w[i]);

                 if(dp[j]<dp[j-w[i]]+w[i]){

                    dp[j]=dp[j-w[i]]+w[i];

                    vis[i][j]=true;

                 }

            }

        }

        for(int i=,j=dp[m]; i<=n && j>; i++){

            if(vis[i][j]){

                cout<<w[i]<<' ';

                 j -= w[i];

            }

        }

        cout<<"sum:"<<dp[m]<<endl;

    }

}

/*

5 3 1 3 4

10 4 9 8 4 2

20 4 10 5 7 4

90 8 10 23 1 2 3 4 5 7

45 8 4 10 44 43 12 9 8 2

*/

一维逆序

#include<cstdio>

#include<string>

#include<cstdlib>

#include<cmath>

#include<iostream>

#include<cstring>

#include<set>

#include<queue>

#include<algorithm>

#include<vector>

#include<map>

#include<cctype>

#include<stack>

#include<sstream>

#include<list>

#include<assert.h>

#include<bitset>

#include<numeric>

#define debug() puts("++++")

#define gcd(a,b) __gcd(a,b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a,b,sizeof(a))

#define sz size()

#define be begin()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

#define all 1,n,1

#define rep(i,n,x) for(int i=(x); i<(n); i++)

#define in freopen("in.in","r",stdin)

#define out freopen("out.out","w",stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int,int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e18;

const int maxn =  ;

const int maxm = 1e5 + ;

const double PI = acos(-1.0);

const double eps = 1e-;

const int dx[] = {-,,,,,,-,-};

const int dy[] = {,,,-,,-,,-};

const int mon[] = {, , , , , , , , , , , , };

const int monn[] = {, , , , , , , , , , , , };

/*

5 3 1 3 4

10 4 9 8 4 2

20 4 10 5 7 4

90 8 10 23 1 2 3 4 5 7

45 8 4 10 44 43 12 9 8 2

*/

int dp[][maxm];

int w[],n,m;

int main()

{

    while(cin>>m>>n)

    {

        memset(dp,,sizeof(dp));

        for(int i=;i<=n;i++) cin>>w[i];

        for(int i=;i<=n;i++)

        {

            for(int j=;j<=m;j++) //注意是顺推

            {

                if(j>=w[i]) dp[i][j]=max(dp[i-][j], dp[i-][j-w[i]]+w[i]);

                else    dp[i][j]=dp[i-][j]; //必须判断

            }

        }

        for(int i=n,j=m; i>; i--)

        {

            if(dp[i][j]>dp[i-][j]) //在顺推时考察第i是放还是不放是从前一个状态推出来的，如果放入i时大，说明放入了i，否则没放和前一个状态一样

            {

               cout << w[i] <<' ';

                j -= w[i];

            }

        }

        cout << "sum:"<<dp[n][m]<<endl;

    }

}

二维顺推

#include<cstdio>

#include<string>

#include<cstdlib>

#include<cmath>

#include<iostream>

#include<cstring>

#include<set>

#include<queue>

#include<algorithm>

#include<vector>

#include<map>

#include<cctype>

#include<stack>

#include<sstream>

#include<list>

#include<assert.h>

#include<bitset>

#include<numeric>

#define debug() puts("++++")

#define gcd(a,b) __gcd(a,b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a,b,sizeof(a))

#define sz size()

#define be begin()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

#define all 1,n,1

#define rep(i,n,x) for(int i=(x); i<(n); i++)

#define in freopen("in.in","r",stdin)

#define out freopen("out.out","w",stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int,int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e18;

const int maxn =  ;

const int maxm = 1e5 + ;

const double PI = acos(-1.0);

const double eps = 1e-;

const int dx[] = {-,,,,,,-,-};

const int dy[] = {,,,-,,-,,-};

const int mon[] = {, , , , , , , , , , , , };

const int monn[] = {, , , , , , , , , , , , };

/*

5 3 1 3 4

10 4 9 8 4 2

20 4 10 5 7 4

90 8 10 23 1 2 3 4 5 7

45 8 4 10 44 43 12 9 8 2

*/

int dp[maxm],vis[maxm];

int w[],n,m;

int main()

{

    while(cin>>m>>n)

    {

        memset(dp,,sizeof(dp));

        memset(vis,,sizeof(vis));

        for(int i=;i<=n;i++) cin>>w[i];

        for(int i=;i<=n;i++)

        {

            for(int j=m;j>=w[i];j--)

            {

                if(dp[j] < dp[j-w[i]]+w[i])

                {

                    dp[j] = dp[j-w[i]]+w[i];

                    vis[j] = vis[j-w[i]] | (<<i);

                }

            }

        }

        for(int i=; i<=n; i++)

        {

            if(vis[m] & (<<i))

            {

                cout << w[i] << ' ';

            }

        }

        cout << "sum:"<<dp[m]<<endl;

    }

}

一维状态压缩

#include<cstdio>

#include<string>

#include<cstdlib>

#include<cmath>

#include<iostream>

#include<cstring>

#include<set>

#include<queue>

#include<algorithm>

#include<vector>

#include<map>

#include<cctype>

#include<stack>

#include<sstream>

#include<list>

#include<assert.h>

#include<bitset>

#include<numeric>

#define debug() puts("++++")

#define gcd(a,b) __gcd(a,b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a,b,sizeof(a))

#define sz size()

#define be begin()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

#define all 1,n,1

#define rep(i,n,x) for(int i=(x); i<(n); i++)

#define in freopen("in.in","r",stdin)

#define out freopen("out.out","w",stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int,int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e18;

const int maxn =  ;

const int maxm = 1e5 + ;

const double PI = acos(-1.0);

const double eps = 1e-;

const int dx[] = {-,,,,,,-,-};

const int dy[] = {,,,-,,-,,-};

const int mon[] = {, , , , , , , , , , , , };

const int monn[] = {, , , , , , , , , , , , };

/*

5 3 1 3 4

10 4 9 8 4 2

20 4 10 5 7 4

90 8 10 23 1 2 3 4 5 7

45 8 4 10 44 43 12 9 8 2

*/

int dp[maxm],vis[maxm];

int w[],n,m;

void print(int i)

{

    if(vis[i]){

        print(vis[i]);

        cout<<i-vis[i]<<' ';

    }

    else{

        cout<<i<<' ';

        return ;

    }

}

int main()

{

    while(cin>>m>>n)

    {

        memset(dp,,sizeof(dp));

        memset(vis,,sizeof(vis));

        for(int i=;i<=n;i++) cin>>w[i];

        for(int i=;i<=n;i++)

        {

            for(int j=m;j>=w[i];j--)

            {

                if(dp[j] < dp[j-w[i]]+w[i])

                {

                    dp[j] = dp[j-w[i]]+w[i];

                    vis[j] = j - w[i];

                }

            }

        }

        print(dp[m]);

        cout << "sum:"<<dp[m]<<endl;

    }

}

一维递归

UVA 624 CD【01背包+路径记录】的更多相关文章

  1. UVA 624 ---CD 01背包路径输出

    DescriptionCD You have a long drive by car ahead. You have a tape recorder, but unfortunately your b ...

  2. UVA 624 - CD (01背包 + 打印物品)

    题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem ...

  3. UVA 624 CD (01背包)

    //路径记录方法:若是dp[j-value[i]]+value[i]>dp[j]说明拿了这个东西,标志为1, //for循环标志,发现是1,就打印出来,并把背包的容量减少,再在次容量中寻找标志: ...

  4. uva 624 CD 01背包打印路径

    // 集训最终開始了.来到水题先 #include <cstdio> #include <cstring> #include <algorithm> #includ ...

  5. UVA--624 CD(01背包+路径输出)

    题目http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...

  6. PAT L3-001 凑零钱(01背包dp记录路径)

    韩梅梅喜欢满宇宙到处逛街.现在她逛到了一家火星店里,发现这家店有个特别的规矩:你可以用任何星球的硬币付钱,但是绝不找零,当然也不能欠债.韩梅梅手边有104枚来自各个星球的硬币,需要请你帮她盘算一下,是 ...

  7. uva 624 CD （01背包）

      CD  You have a long drive by car ahead. You have a tape recorder, but unfortunately your best musi ...

  8. UVA 624 CD（01背包+输出方案）

    01背包,由于要输出方案,所以还要在dp的同时,保存一下路径. #include <iostream> #include <stdio.h> #include <stri ...

  9. UVA 624 CD(DP + 01背包)

    CD You have a long drive by car ahead. You have a tape recorder, but unfortunately your best music i ...

随机推荐

  1. 剑指Offer - 九度1362 - 左旋转字符串（Move!Move!!Move!!!）

    剑指Offer - 九度1362 - 左旋转字符串(Move!Move!!Move!!!)2013-11-23 03:05 题目描述: 汇编语言中有一种移位指令叫做循环左移(ROL),现在有个简单的任 ...

  2. PHP vscode+XDebug 远程断点调试服务器上的代码

    对于简单的项目或仅仅想知道某一位置的某个变量是什么值,直接使用var_dump配置exit来打印和中断就可以了,方便又快捷, 而对于大型项目的调试,或想了解某个系统的整个运行过程,xdebug可能会是 ...

  3. NodeJs03 express框架 Todo商城

    前言 由于NodeJs本身的异步非阻塞特性和对http的天然支持,所以使用NodeJs编写高性能,可伸缩的Web服务器非常简单.开发完整的Web服务器还需要路由,错误处理,请求拦截,请求和响应的解析, ...

  4. Spring 笔记（一）概念梳理

    概念 预备知识 1. POJO POJO是Plain Old Java Object的缩写,是软件开发大师Martin Fowler提出的一个概念,指的是一个普通Java类.也就说,你随便编写一个Ja ...

  5. JavaScript里面的正则以及eval

    1.eval JavaScript中的eval是Python中eval和exec的合集,既可以编译代码也可以获取返回值. eval() EvalError   执行字符串中的JavaScript代码 ...

  6. google protobuf 中的proto文件编写规则

    1. 简单介绍 protobuf文件:就是定义你要的消息(类似Java中的类)和消息中的各个字段及其数据类型(类似java类中的成员变量和他的数据类型) 2. Protobuf消息定义 消息由至少一个 ...

  7. BZOJ 2438：杀人游戏（tarjan+概率）

    杀人游戏Description一位冷血的杀手潜入 Na-wiat,并假装成平民.警察希望能在 N 个人里面,查出谁是杀手. 警察能够对每一个人进行查证,假如查证的对象是平民,他会告诉警察,他认识的人, ...

  8. MPSVPX 配置

    MPSVPX 配置 设置主机名,IP地址,掩码,网关,DNS服务器,时区(使用WebGUI界面设置). bash-2.05b# cat svm.conf arp -d -a route flush i ...

  9. GYM - 101147 B.Street

    题意: 大矩形代表市场,大矩形当中有很多小矩形样式的伞.这些小矩形都贴着大矩形的左边或者右边且互不相交.小矩形以外的地方都是阳光.求经过大矩形时在阳光下的最短时间. 题解: 最短路的做法.起点和终点与 ...

  10. SPOJ - BALNUM Balanced Numbers

    题意: 求出所给范围内满足其数位上的奇数出现偶数次,数位上的偶数出现奇数次(或不出现)的数的个数. 思路: 对于0 ~ 9 每个数有3种情况. 1.没出现过 2.出现奇数次 3.出现偶数次 那么就可以 ...