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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 

Output 

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

解题思路：path[i][j]记录策略的选择，输出的时候倒推回去

#include <iostream>
#include <cstdio>
#include <cstring>
#define N 10005
using namespace std;
int main()
{
	int m,n,i,j;
	int dp[N],path[25][N],v[25];
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	while(~scanf("%d%d",&n,&m))
	{
		for(i=0;i<m;i++)
			scanf("%d",v+i);
		memset(dp,0,sizeof(dp));
		memset(path,0,sizeof(path));
		for(i=0;i<m;i++)
			for(j=n;j>=v[i];j--)
				if(dp[j]<dp[j-v[i]]+v[i])
				{
					dp[j]=dp[j-v[i]]+v[i];
					path[i][j]=1;
				}
				else
					path[i][j]=0;
		j=n;
		for(i=m-1;i>0;i--)
		{
			if(path[i][j]==1 && j>=0)  
            {  
                printf("%d ",v[i]);  
                j-=v[i];  
            }  
		}
		printf("sum:%d\n",dp[n]);
	}
    return 0;
}

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标签：style   blog   color   strong   os   2014   

原文地址：http://blog.csdn.net/hqu_fritz/article/details/36446889