UVA 624 CD （01背包+打印路径）

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

０１背包＋打印路径，用数组ｖｉｓ［ｉ］［ｊ］来表示ｉ是否取。输出方案的时候要注意：从N到1输出时，如果f[i][v]==f[i-1][i-v]及f[i][v]==f[i-1][f-c[i]]+w[i]同时成立，应该按照后者（即选择了物品i）来输出方案。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[1100][1100],vis[1100][1100],a[1100];
int main()
{
int sum,n,i,j,k,l;
while(scanf("%d%d",&sum,&n)!=EOF)
{
memset(vis,0,sizeof(vis));
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
for(i=1;i<=n;i++)
for(j=0;j<=sum;j++)//01背包状态转移方程：
//dp[i][j]=max(dp[i-1][j],dp[i-1][j-a[i]]+a[i])
{
dp[i][j]=dp[i-1][j];
if(j>=a[i])
{
if(dp[i][j]<dp[i-1][j-a[i]]+a[i])
{
vis[i][j]=1;//表示物品是否取
dp[i][j]=dp[i-1][j-a[i]]+a[i];
}
}

}
int k=sum;
for(i=n;i>=1;i--)
{
if(vis[i][k]==1&&k>=0)
{
printf("%d ",a[i]);
k=k-a[i];
}

}
printf("sum:%d\n",dp
[sum]);
}
return 0;
}