HDU 2602 bone collector (01 backpack problem)

Bone Collector

Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 2080 accepted submission (s): 669

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 the 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

Http://acm.hdu.edu.cn/showproblem.php? PID = 2602 detailed algorithm description see http://baike.baidu.com/view/1731915.htm? Fr = ala0 the simplest Recursion Method: (this algorithm will time out .) // The time complexity is 2 ^ n </P> <p> # include <iostream> <br/> # include <cmath> <br/> # include <cstring> <br /># include <map> <br/> # include <algorithm> <br/> using namespace STD; </P> <p> const int max = 1001; <br/> typedef struct <br/>{< br/> int value; <br/> int volume; <br/>} bone; </P> <p> bone B [Max]; </P> <p> int make (int I, Int J) // item I, current remaining capacity j <br/>{< br/> if (I <= 0) <br/> return 0; <br/> int R1 = 0, R2 = 0; <br/> If (j> = B [I]. volume) // The value of the I-th item that can be placed in a backpack <br/>{< br/> R1 = make (I-1, J-B [I]. volume) + B [I]. value; <br/>}< br/> r2 = make (I-1, J ); // The value of the I-th item that is not included in the backpack <br/> If (r1> R2) <br/> return R1; <br/> else <br/> return R2; <br/>}</P> <p> int main () <br/>{< br/> freopen ("in6.txt", "r", stdin); <br/> int t, n, V; <br/> CIN> T; <br/> while (t --) <br/> {<br/> CIN> N> V; <br/> for (INT I = 1; I <= N; I ++) <br/> {<br/> CIN> B [I]. value; <br/>}< br/> for (INT I = 1; I <= N; I ++) <br/>{< br/> CIN> B [I]. volume; <br/>}< br/> int T = make (n, V); <br/> cout <t <Endl; <br/>}</P> <p>Use Dynamic Planning. The time complexity is N * V (Number * capacity), and the space complexity two-dimensional matrix (N * V) N * V may be greater than 2 ^ n # Include <iostream> <br/> # include <cmath> <br/> # include <cstring> <br/> # include <map> <br/> # include <Algorithm> <br/> using namespace STD; </P> <p> const int max = 1001; <br/> typedef struct <br/>{< br/> int value; <br/> int volume; <br/>} bone; </P> <p> bone B [Max]; </P> <p> int f [Max] [Max]; </P> <p> int main () <br/> {<br/> int t, n, V; <br/> CIN> T; <br/> while (t --) <br/>{< br/> CIN> N> V; <br/> for (INT I = 1; I <= N; I ++) <br/>{< br/> CIN> B [I]. value; <br/>}< br/> for (INT I = 1; I <= N; I ++) <br/>{< br/> CIN> B [I]. volume; <br/>}< br/> for (INT I = 0; I <= V; I ++) <br/>{< br/> F [0] [I] = 0; <br/>}< br/> for (INT I = 1; I <= N; I ++) <br/>{< br/> for (Int J = 0; j <= V; j ++) <br/> {<br/> F [I] [J] = f [I-1] [J]; <br/> If (j> = B [I]. volume & F [I-1] [J-B [I]. volume] + B [I]. value> F [I] [J]) <br/> F [I] [J] = f [I-1] [J-B [I]. volume] + B [I]. value; <br/>}< br/> cout <F [N] [v] <Endl; <br/>}< br/>}Simplified matrix algorithm. The time complexity is also N * V, and the space complexity is a one-dimensional array (V ). # Include <iostream> <br/> # include <cmath> <br/> # include <cstring> <br/> # include <map> <br/> # include <Algorithm> <br/> using namespace STD; </P> <p> const int max = 1090; <br/> typedef struct <br/>{< br/> int value; <br/> int volume; <br/>} bone; </P> <p> bone B [Max]; <br/> int DP [Max]; </P> <p> int main () <br/> {<br/> // freopen ("in6.txt", "r", stdin ); <br/> int t, n, V; <br/> CIN> T; <br/> while (t --) <br/>{< br/> memset (DP, 0, sizeof (DP); <br/> CIN> N> V; <br/> for (INT I = 1; I <= N; I ++) <br/> {<br/> CIN> B [I]. value; <br/>}< br/> for (INT I = 1; I <= N; I ++) <br/>{< br/> CIN> B [I]. volume; <br/>}< br/> for (INT I = 1; I <= N; I ++) <br/>{< br/> for (Int J = V; j> = B [I]. volume; j --) <br/>{< br/> If (DP [J] <DP [J-B [I]. volume] + B [I]. value) <br/>{< br/> DP [J] = DP [J-B [I]. volume] + B [I]. value; <br/>}< br/> cout <DP [v] <Endl; <br/>}< br/> return 0; <br/>}