uva 624 CD (01背包+打印路径)_分组背包打印路径问题-优快云博客

本文链接：https://blog.youkuaiyun.com/sdjzping/article/details/11176989

本文介绍了一个经典的01背包问题案例，并详细展示了如何通过定义一个额外的二维数组来记录选择过程，以便最终能够输出所选物品及其总价值。通过对具体样例的解析，帮助读者理解算法实现。

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1、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=565

2、01背包问题，注意打印路径的方式，可以定义一个二维的path数组，记录是否选中此物品，

3、题目：

CD

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 20
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 =5, number of tracks=3, first track lasts for 1 minute, second one 3 minutes, next one 4 minutes

Output

Set of tracks (and durations) which are the correct solutions and string `` sum: " and sum of duration times.

Sample Input

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

Sample Output

1 4 sum:5
8 2 sum:10
10 5 4 sum:19
10 23 1 2 3 4 5 7 sum:55
4 10 12 9 8 2 sum:45

4、AC代码：

#include<stdio.h>
#define N 10000
#include<algorithm>
using namespace std;
int w[N];
int dp[N][N];
int path[N][N];
int main()
{
    int v,n;
    while(scanf("%d%d",&v,&n)!=EOF)
    {
        for(int i=1; i<=n; i++)
            scanf("%d",&w[i]);
        for(int i=0; i<=n; i++)
        {
            for(int j=0; j<=v; j++)
                dp[i][j]=0,path[i][j]=0;
        }
        for(int i=1; i<=n; i++)
        {
            for(int j=0; j<=v; j++)
            {
                dp[i][j]=dp[i-1][j];
                if(j>=w[i])
                {
                    //dp[i][j]=max(dp[i][j],dp[i-1][j-w[i]]+w[i]);
                    if(dp[i][j]<dp[i-1][j-w[i]]+w[i])
                    {
                        path[i][j]=1;
                        dp[i][j]=dp[i-1][j-w[i]]+w[i];
                    }
                    //printf("*%d %d* ",i,j);
                }
            }
        }
        int j=v;
        for(int i=n; i>=1; i--)
        {
            //printf("&%d %d& ",w[i],j);
            if(path[i][j]==1 && j>=0)
            {
                printf("%d ",w[i]);
                j-=w[i];
            }
        }
        printf("sum:%d\n",dp[n][v]);
    }
    return 0;
}