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 20no track is longer than N minutestracks do not repeatlength of each track is expressed as an integer numberN 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

题目大意：每行就是一组数据，每组数据前两个数n， m代表，目标数和拥有的数据数，接下来有m个数据。要求由m个数据可以组成的最接近目标数的数。输出该数，和组成该数的数。（每个数只能用一次）。

解题思路：01背包。要注意每个数只能用一次。并且答案不唯一。

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>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;}