uva 624 CD

01背包写成一维的优点是省内存，坏处是中间过程都没被保存，而这题要求打出路径

假设直接写一维的然后标记，是不是有些本末倒置

假设写记忆化搜索，我没想出来怎么写……

也懒得想了。反正有现成的二维形式，二维的优点就是保存了路径

所以非常easy能够回溯出路径，事实上这里假设写成一维的反而浪费了内存，由于你还要

多开一个二维标记。所以说是本末倒置

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int dp[30][10000];
int a[30];
int n,m;
int main()
{
	int i,j;
	while(cin>>n>>m)
	{
		memset(dp,0,sizeof(dp));
		for(i=1;i<=m;i++)
			cin>>a[i];
		for(i=1;i<=m;i++)
			for(j=n;j>=0;j--)
				if(j>=a[i]&&dp[i-1][j]<dp[i-1][j-a[i]]+a[i])
					dp[i][j]=dp[i-1][j-a[i]]+a[i];
				else
					dp[i][j]=dp[i-1][j];
		i=m;
		j=n;
		while(i>0)
		{
			if(j>=a[i]&&dp[i][j]==dp[i-1][j-a[i]]+a[i])
			{
				cout<<a[i]<<" ";
				j-=a[i];
			}
			i--;
		}
		printf("sum:%d
",dp[m][n]);
	}
	return 0;
}

  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