CD UVA - 624 ————01背包（寻找路径）

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

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这题让找的是01背包的路径，是一个很好的题，能够加深自己对背包问题的转移过程的理解，
寻找背包路径的过程的重点代码：

        for(int i=n-1;i>=0;i--)
            for(int j=W;j>=v[i];j--)
                if(dp[j-v[i]]+v[i]>dp[j])
                {
                    dp[j]=dp[j-v[i]]+v[i];//这道题价值和重量是一样的
                    vis[i][j]=1;    
                }
        int j=W;//刚开始是最大重量的时候
        for(int i=0;i<n;i++)
        {
            if(vis[i][j])//如果物品被放入背包
            {
                printf("%d ",v[i]);//按照正序输出（拿出）背包里物品的重量/价值
                j -= v[i];  //让背包减去拿出去的物品
            }   

---

ACcode：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=100000+7;
int dp[MAXN];
bool vis[25][MAXN];
int n,W,v[MAXN];
int main()
{
    while(~scanf("%d %d",&W,&n))
    {
        for(int i=0;i<n;i++)
            scanf("%d",&v[i]);
        memset(dp,0,sizeof(dp));
        memset(vis,0,sizeof(vis));
        for(int i=n-1;i>=0;i--)
            for(int j=W;j>=v[i];j--)
                if(dp[j-v[i]]+v[i]>dp[j])
                {
                    dp[j]=dp[j-v[i]]+v[i];
                    vis[i][j]=1;    
                }
        int j=W;
        for(int i=0;i<n;i++)
        {
            if(vis[i][j])
            {
                printf("%d ",v[i]);
                j -= v[i];  
            }   
        } 
        printf("sum:%d\n",dp[W]);

    }
    return 0;
}