杭电ACM 2602 Bone Collector背包

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 58027 Accepted Submission(s): 24207

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

Author

Teddy

Source

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

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做的第一道01背包题。

题意：骨头收集者！！骨头都有自己的value（价值）和volume（体积），现在给一个一定容积的袋子，问如何装才能装下最大价值的骨头。典型的01背包问题。以下列程序为例，若以二维数组d[1005][1005]来记录01背包表的值的话，动态方程为d[i][j]=max(d[I-1][j],d[I-1][j-b[I]]+a[I])以题目给定例子来画出下面的01背包表：

该表的第一行表示背包的大小j，第一列表示物品的价值a[I]，这个表造成的疑问在于，列如为什么d[1][9]不是价值为1的物品加上价值为2的物品，他们的体积相加刚好是9啊！？这是因为在I=1这行，没有轮到价值大于1的物品来排。也就是说，第I行就是价值为1~I的物品去进行装填。按照这个道理，右下角最后一个格子里的就是我们要求的最大值。或许还是没懂，其实自己亲自把表里的值推推就会好些了。

原理懂了之后编程是水到渠成啊！

附上代码：

#include<iostream>#include<cmath>#include<cstring>using namespace std;int a[1005],b[1005];int d[1005][1005];int main(){    int n;    cin>>n;    while(n--)    {        memset(d,0,sizeof(d));        int x,y;        cin>>x>>y;        for(int i=1;i<=x;i++)            cin>>a[i];///价值        for(int i=1;i<=x;i++)            cin>>b[i];///所占空间        for(int i=1;i<=x;i++)///动态规划        {            for(int j=0;j<=y;j++)            {                if(j>=b[i])                    d[i][j]=d[i-1][j]>(d[i-1][j-b[i]]+a[i])?d[i-1][j]:(d[i-1][j-b[i]]+a[i]);                else                    d[i][j]=d[i-1][j];            }        }        cout<<d[x][y]<<endl;    }    return 0;}

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