UVA 624 – CD 解题报告

1.原题:

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

2.中文大意

就是给定规定的时间N，然后给M个磁带，每个磁带的时间都告诉你了，然后求不大于时间N的总时间是多少，将其序列和最接近N的总时间输出

3.思路

思路很简单了，就是用一个vis的数组先进行暴力枚举，将所有的时间和求出，然后每次算时间的时候，与N进行比较，再用一个int来存储与N的距离，即N-Sum，然后每次都进行判断，看是否小于N-Sum，这样就可以求出来了

4.代码

include<iostream>
include<cstring>
using namespace std;

int sum = 0;
int tapeNum = 0;
int numList[200];
int vis[200];
int resultSet[200];
bool qq = false;
int dis = 100000;
int nn = 0;

void checkSum(int index){
    int op = 0;
    for (int i = 0;i <= index;i++){
        if(vis[i])
            op += numList[i];
    }
    if(sum >= op) {
        if(sum - op < dis){
            if(sum - op == 0) qq = true;
            int pp = 0;
            dis = sum - op;
            for (int i = 0;i<=index;i++) {
                if(vis[i])
                    resultSet[pp++] = numList[i];
            }
            nn = pp;
        }
    }
}

void getResult(int index){
    if(qq || index == tapeNum) return ;
    vis[index] = 1;
    checkSum(index);
    getResult(index+1);
    vis[index] = 0;
    checkSum(index);
    getResult(index+1);
}

int main(){

    while(cin>>sum>>tapeNum) {
        memset(numList,0,sizeof(numList));
        memset(vis,0,sizeof(vis));
        memset(resultSet,0,sizeof(resultSet));
        qq = false;
        nn = 0;
        dis = 100000;
        for (int i = 0 ;i < tapeNum ;i ++) cin >> numList[i];
        getResult(0);
        for (int i = 0;i < nn;i++) {
            cout<<resultSet[i]<<" ";
        }
        cout<<"sum:"<<sum - dis<<endl;
    }
    return 0;
}