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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：求和最大的，弄了两个小时没鼓捣出来=_=
两种方法：一种是二维的dp[i][j] 表示前 i 种体积为 j 的最大价值量，那么对于 dp[i][j] = dp[i - 1][j]就一定不输出咯，因为这个表示第i个没选
第二种对于一维的来说：dp[i]表示 体积 为 i 的最大价值量，然后可以设一个数组来记录
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <cstdio> 5 using namespace std; 6 const int Max = 1000 + 10; 7 int trak[Max], dp[Max * 20]; 8 int vis[Max * 20][Max]; 9 int vistrak[Max]; 10 int main() 11 { 12 int n; 13 while (scanf("%d", &n) != EOF) 14 { 15 int traks; 16 scanf("%d", &traks); 17 for (int i = 1; i <= traks; i++) 18 scanf("%d", &trak[i]); 19 memset(vis, 0, sizeof(vis)); 20 memset(dp, 0, sizeof(dp)); 21 memset(vistrak, 0, sizeof(vistrak)); 22 for (int i = 1; i <= traks; i++) 23 { 24 for (int j = n; j >= trak[i]; j--) 25 { 26 if (dp[j] <= dp[j - trak[i]] + trak[i]) 27 { 28 dp[j] = dp[j - trak[i]] + trak[i]; 29 vis[j][i] = trak[i]; // 体积为 j 时选 i 的价值 30 } 31 } 32 } 33 int tempn = n; 34 for (int i = traks; i > 0; i--) 35 { 36 if (vis[tempn][i]) // 如果选了就应该是输出的， 37 { 38 vistrak[i] = 1; 39 tempn -= vis[tempn][i]; 40 } 41 } 42 for (int i = 1; i <= traks; i++) 43 { 44 if (vistrak[i]) 45 printf("%d ", trak[i]); 46 } 47 printf("sum:%d\n", dp[n]); 48 } 49 return 0; 50 }
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原文地址：http://www.cnblogs.com/zhaopAC/p/5517996.html
