HDU 2602 bone collector (0/1 backpack problem)

Bone Collector

Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 22041 accepted submission (s): 8912

Problem descriptionpolicyears ago, in Teddy's hometown there was a man who was called "Bone Collector ". this man like to collect varies of bones, such as dog's, cow's, also he went to the grave...
The bone collector had a big bag with a volume of V, and along his trip of collecting there are a lot of bones, obviusly, different bone has different value and different volume, now given the each bone's value along his trip, Can you calculate out
Maximum Of the total value the bone collector can get?

 

Inputthe first line contain a integer t, the number of instances.
Followed by T cases, each case three lines, the first line contain two integer N, V, (n <= 1000, v <= 1000) representing the number of bones and the volume of his bag. and the second line contain N integers representing the value of each bone. the third
Line contain N integers representing the volume of each bone.

Outputone integer per line representing the maximum of the total value (This number will be less than 231 ).

Sample Input

15 101 2 3 4 55 4 3 2 1

 

Sample output

14

Question: give you the value of N bones and the corresponding volume. In a provided V-capacity backpack, find the maximum value that can be loaded.

 

Method 1: One-dimensional array

 

Import Java. io. *; import Java. util. *; public class main {int t, n, V, M = 1005; int DP [] = new int [m]; public static void main (string [] ARGs) {New Main (). work ();} void work () {vertex SC = new vertex (New bufferedinputstream (system. in); t = SC. nextint (); While (t --! = 0) {n = SC. nextint (); V = SC. nextint (); node [] = new node [N]; arrays. fill (DP, 0); For (INT I = 0; I <n; I ++) {node [I] = new node (); node [I]. N = SC. nextint () ;}for (INT I = 0; I <n; I ++) {node [I]. V = SC. nextint () ;}for (INT I = 1; I <= N; I ++) {for (Int J = V; j> = 0; j --) {If (j> = node [I-1]. v) {int A = node [I-1]. v; // the volume of each bone int B = node [I-1]. n; // The value of each bone DP [J] = math. max (DP [J], DP [J-A] + B); // obtain the maximum value of each bone} system. out. println (DP [v]) ;}} class node {int N; int v ;}}

Method 2: Two-dimensional array

Import Java. io. *; import Java. util. *; public class main {int N, V, T, M = 1005; int [] [] value; public static void main (string [] ARGs) {New Main (). work ();} void work () {vertex SC = new vertex (New bufferedinputstream (system. in); t = SC. nextint (); While (t --! = 0) {n = SC. nextint (); V = SC. nextint (); node [] = new node [N]; value = new int [m] [m]; for (INT I = 0; I <N; I ++) {node [I] = new node (); node [I]. N = SC. nextint () ;}for (INT I = 0; I <n; I ++) {node [I]. V = SC. nextint () ;}for (INT I = 1; I <= N; I ++) {for (Int J = 0; j <= V; j ++) {If (j <node [I-1]. v) value [I] [J] = value [I-1] [J]; else {int A = node [I-1]. n; // The value of each bone int B = node [I-1]. v; // the volume of each bone. value [I] [J] = math. max (value [I-1] [J], A + value [I-1] [J-B]); // obtain the maximum value of each bone} system. out. println (value [N] [v]) ;}} class node {int N; int v ;}}