uva 624 CD （01背包）

uva 624 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

題目大意：每行就是一組數據，每組數據前兩個數n， m代表，目標數和擁有的數據數，接下來有m個數據。要求由m個數據可以組成的最接近目標數的數。輸出該數，和組成該數的數。（每個數只能用一次）。

解題思路：01背包。要注意每個數只能用一次。並且答案不唯一。

#include
#include
#include
#include
typedef long long ll;
using namespace std;
int num[25];
int dp[100005], rec[100005], r[25];
int main() {
    int n, m, cnt;
    while (scanf("%d %d", &n, &m) == 2) {
        cnt = 0;
        memset(num, 0, sizeof(num));
        memset(dp, 0, sizeof(dp));
        memset(rec, 0, sizeof(rec));
        for (int i = 0; i < m; i++) {
            scanf("%d", &num[i]);
        }
        dp[0] = 1;
        for (int i = 0; i < m; i++) {
            for (int j = n; j >= 0; j--) {
                if (dp[j] && !dp[j + num[i]]) {
                    dp[j + num[i]] = 1;
                    rec[j + num[i]] = num[i];
                }
            }
        }

        while (!dp[n]) n--;
        int t = n;
        while (n >= 0 && rec[n]) {
            r[cnt++] = rec[n];
            n -= rec[n];
        }
        for (int i = cnt - 1; i >= 0; i--) {
            printf("%d ", r[i]);
        }
        printf("sum:%d\n", t);
    }
    return 0;
}