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
思路：
本题是典型的01背包问题，但要考虑记忆化搜索，对选择的物品要用一个数组来储存他是否选择。
代码
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,T,t[100000],dp[30][100000],ap[30][100000]={0},a[100000];
ap表示是否选择该物品
while(scanf("%d%d",&T,&n)!=EOF){
for(int i=1;i<=n;i++)cin>>t[i];
for(int i=1;i<n;i++)dp[i][0]=0;//初始化
for(int j=0;j<=T;j++){
dp[0][j]=0;
ap[0][j]=1;//初始化0件物品时，容量为j值为1;
}
for(int i=1;i<=n;i++){
for(int j=1;j<=T;j++){
ap[i][j]=0;
if(j<t[i])dp[i][j]=dp[i-1][j];
else{
dp[i][j]=max(dp[i-1][j],dp[i-1][j-t[i]]+t[i]);
if(dp[i-1][j]!=dp[i][j])ap[i][j]=1;//选择第i件物品
}
}
}
int maxvalue=dp[n][T];
int cnt=0;
for(int i=n,j=T;i>=1;i--){
if(ap[i][j]==1){
a[cnt]=t[i];
cnt++;
j-=t[i];
}
}//从容量为T开始找，如果ap为1则减掉该物品的容积。
for(int i=cnt-1;i>=0;i--)printf("%d ",a[i]);
printf("sum:%d\n",maxvalue);
}
return 0;
}
