Hdoj 2602 Bone collector--01 Backpack

Source of the topic:HPU 2602-- Bone Collector

Problem Description

Manyyears ago, in Teddy's hometown there was a man who was called "Bonecollector". 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 he trip ofcollecting there is a lot of bones, obviously , different bone have differentvalue and different volume, now given the each bone's value along his trips, can you Calculat E out the maximum of the total value the bone collector can get?

Input

Thefirst line contain a integer T, the number of cases.
Followed by T cases, each case three lines, the first line contain Twointeger N, V, (N <=, v <=) repre Senting the number of Bonesand the volume of his bag. And the second line contain N integers representingthe value of each bone. The third line contain N integers representing thevolume of each bone.

Output

Oneinteger per line representing the maximum of the total value (this number willbe less than 231).

Sample Input

1

5 10

1 2 3) 4 5

5 4 3) 2 1

Sample Output

14

 

Research point: Dynamic Planning-01 Backpack

Main topic:

Given the T-Group test data, each set of data is given an n (a bone) and a V (the capacity of the Backpack V), and the next two lines are given the value and volume of the N-block bone respectively. The maximum value that can be loaded by the output backpack.

Topic Analysis:

dynamic Programming equation: F (n,m) =max{f (N-1,m), F (n-1,m-w[n]) +v (n,m)}, according to this recursion, we know that the 01 backpack records the optimal solution of the current position.

AC Code:

#include <cstdio> #include <cstring> #include <cstring> #include <algorithm>using namespace std int value[1005],volume[1005];//bone value, Volume int dp[1005][1005];//record Backpack volume value from 0 to V max value int main () {int t;int n,v;scanf ("%d", &t), while (t--) {scanf ("%d%d", &n,&v), memset (Dp,0,sizeof (DP)),//DP array initialized to 0 for (int i=1;i<=n;i++) { scanf ("%d", &value[i]);} for (int i=1;i<=n;i++) {scanf ("%d", &volume[i]);} for (int i=1;i<=n;i++) {for (int j=0;j<=v;j++) {if (J<volume[i])//Current Backpack volume J Cannot load bones of volume Volume[i] {dp[i][j]=dp[i-1 ][J]; }else//Current Backpack Volume J can pack a bone of volume Volume[i] {//select Add or not add value for current backpack Dp[i][j]=max (Dp[i-1][j],dp[i-1][j-volume[i]]+value[i]);}}} printf ("%d\n", Dp[n][v]);} return 0;}

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Hdoj 2602 Bone collector--01 Backpack