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JXUST第二赛-B. Bone Collector

2024-07-13 04:33:12   216人阅读

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

 1 /************************************************************************* 2     > File Name: 0-1.cpp 3     > Author: Mercu 4     > Mail: bkackback1993@gmail.com 5     > Created Time: 2014年07月20日 星期日 09时17分15秒 6  ************************************************************************/ 7  8 #include<iostream> 9 #include<string.h>10 using namespace std;11 12 int t[1005],v[1005];13 int f[1005];14 int ozzy(int a,int b)15 {16     17 }18 int main()19 {20     int o,a,i,b,c;21     int maxv;22     cin>>o;23     while(o--)24     {25         int j,k,l;26         cin>>a>>b;27         memset(v,0,sizeof(v));28         memset(t,0,sizeof(t));29         memset(f,0,sizeof(f));30         for(i = 1;i <= a;i++)31         {32             cin>>v[i];33         }34         for(i = 1;i <= a;i++)35         {36             cin>>t[i];37         }38         39         for(i = 1;i <= a;i++)40         {41             for(j = b;j >= t[i];j--)42             {43                 if(t[i] <= j)44                 f[j] = max(f[j],f[j - t[i]] + v[i]);45             }46         }47         cout<<f[b]<<endl;48     }49     return 0;50 }

　　

　　这个题是一个很明显的0-1背包问题,就是给定的物品,有两种状态,一种是放,一种是不放,一个物品不能放多次.

1         for(i = 1;i <= a;i++)2         {3             for(j = b;j >= t[i];j--)4             {5                 if(t[i] <= j)6                 f[j] = max(f[j],f[j - t[i]] + v[i]);7             }8         }

 

　　这一段是核心代码,要注意for的范围,第一个for从1到a,a是什么?a是物品的个数,第二个for从b到t[i]  t[i]是物品的体积.b是最大体积,所以

 

1 for 1 -->n //  1/0取决于输入w[i] v[i]的时候下标0开始还是1开始2     for b -->w[i]3     if(w[i] <= b)4         f[j] = max(f[j],f[j-t[i]] + v[i]]);
其中,w[i]是质量,v[i]是价值(质量和价值 == 体积和价值).

 

 


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