Algorithm学习笔记 --- Bone Collector

Problem Description

Many years 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 , obviously , 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 ?

 

Input

The first line contain a integer T , the number of cases.
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.

 

Output

One 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

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

解题分析：此题为01背包的简单题型。带入：dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]);

代码：

#include <iostream>
#include <string.h>
#define max(a,b) (a>b)?a:b


#define N 1010
using namespace std;


int value[N], volume[N], dp[N];


// 0-1背包，优化空间
void dpPackage(int n, int v)
{
    int i, j;


    memset(dp, 0, sizeof(dp));


    for (i = 1; i <= n; i ++) {
        for (j = v; j >= volume[i]; j --) {
                dp[j]=max(dp[j],dp[j-volume[i]]+value[i]);//dp[j] = dp[j] > dp[j - volume[i]] + value[i] ? dp[j] : dp[j - volume[i]] + value[i];
        }
    }


    cout<<dp[v];
}


int main( )
{
    int i, T, n, v;


    cin>>T;


    while (T --) {
        // 接收参数
        cin>>n>>v;


        for (i = 1; i <= n; i ++)
            cin>>value[i];
        for (i = 1; i <= n; i ++)
            cin>>volume[i];


        // 0-1背包
        dpPackage(n, v);
    }


    return 0;
}