CD 路径记忆

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

题意：一张磁带长度为n的CD，有m首歌，问在这张CD上最多能刻长度多少的歌。
整首歌要全部被刻。

解题思路：简单的01背包问题，但输出是个重点，需要用到路径记忆。用vis数组标记第i首歌在长度为j时被刻录。最后输出时，需要逆序访问。

Code：

#include<iostream>
#include<cstring>
using namespace std;
int m,n;
const int maxn=1e5+7;
long long dp[maxn];
int vis[25][maxn];
int w[25];

int main(){
	while(cin>>m>>n){
		memset(dp,0,sizeof(dp));
		memset(vis,0,sizeof(vis));
		for(int i=0;i<n;i++) cin>>w[i];
		for(int i=0;i<n;i++){
			for(int j=m;j>=w[i];j--){
				if(dp[j-w[i]]+w[i]>=dp[j]){//判断很重要，不然路径存的位置与所求会不符 
					dp[j]=dp[j-w[i]]+w[i];
					vis[i][j]=1;//路径记忆 
				}						
			}
		}
		int w1[25],q=0;
		for(int i=n-1,j=m;i>=0;i--){
			if(vis[i][j]){
				//cout<<w[i]<<" ";
				w1[q++]=w[i];
				j=j-w[i]; 
			}
		}
		for(int i=q-1;i>=0;i--){//与题目做到相同输出 
			cout<<w1[i]<<" ";
		}
		printf("sum:%lld\n",dp[m]);
	}
	
	return 0;
}