B - CD UVA - 624

B - CD UVA - 624

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 M and durations of the tracks. For example from first line in sample data: N=5, M=3, first track lasts for 1 minute, second one 3 minutes, next one 4 minutes
The input data satisfies the following constraints:
N≤10000
M≤20

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 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

简单来说就是背包问题，多组数据，先输入两个数字，第一个表示最大空间，第二表示磁带个数，后面输入每个磁带空间，从中选出磁带，要求空间尽可能接近最大。
标准背包问题，模板往里套，用二维数组记录所选择的磁带

#include<iostream>
#include<algorithm>
#include<cstring>
#include<stack>
using namespace std;
int main()
{
	stack<int>s1;
	int t=0,n,m,i,j,s[25],dp[10005],a[10005][25];
	while(cin>>n>>m)
	{
		for(i=0;i<m;i++)
			cin >> s[i];
		memset(dp,0,sizeof(dp));
		memset(a,0,sizeof(a));
		for(i=0;i<m;i++)
			for(j=n;j>=s[i];j--)
			if(dp[j]<=dp[j-s[i]]+s[i])
			{
				dp[j]=dp[j-s[i]]+s[i];
				a[j][i]=1;if(!t)t= j;
			}
		for(i=m-1,j=n;i>=0;i--)
		if(a[j][i]){
			s1.push(s[i]);
			j-=s[i];
		}
		while(!s1.empty())
		{
			cout << s1.top()<<" ";
			s1.pop();
		}
		cout << "sum:"<<dp[n]<<endl;
	}
	return 0;
}