HDU 2602 Bone Collector（01背包）

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

【思路分析】

   裸的01背包问题。用了两种方法，第一种是用二维数组，第二种是用滚动数组（一维），两种方法都可以一边输入一边处理。滚动数组的具体操作见小白书。

代码如下：

1、二维数组

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int maxn = 1005;int n,v;int vol;int val[maxn];int dp[maxn][maxn];int maxs(int a,int b){    return a > b ? a : b;}void init(){   scanf("%d %d",&n,&v);   for(int i = 1;i <= n;i++)   {      scanf("%d",&val[i]);   }   for(int i = 1;i <= n;i++)   {       scanf("%d",&vol);       for(int j = 0;j <= v;j++)       {          dp[i][j] = (i == 1 ? 0 : dp[i - 1][j]);          //dp[i][j]表示把前i个物品放到体积为j的背包中的最大总价值          if(j >= vol)          {             dp[i][j] = maxs(dp[i][j],dp[i - 1][j - vol] + val[i]);             //不要该物品以及要该物品两种情况          }       }   }}int main(){    int t;    scanf("%d",&t);    while(t--)    {        init();        printf("%d\n",dp[n][v]);    }    return 0;}

2、滚动数组

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int maxn = 1005;int n,v;int vol;int val[maxn];int dp[maxn];int maxs(int a,int b){    return a > b ? a : b;}void init(){   memset(dp,0,sizeof(dp));   scanf("%d %d",&n,&v);   for(int i = 1;i <= n;i++)   {      scanf("%d",&val[i]);   }   for(int i = 1;i <= n;i++)   {       scanf("%d",&vol);       for(int j = v;j >= 0;j--)       {           if(j >= vol)           {               dp[j] = maxs(dp[j],dp[j - vol] + val[i]);               //滚动数组,dp[j]中相当于保存dp[i - 1][j],dp[j - vol]中相当于保存dp[i - 1][j - vol]           }       }   }}int main(){    int t;    scanf("%d",&t);    while(t--)    {        init();        printf("%d\n",dp[v]);    }    return 0;}