hdu 2602 Bone Collector

简单的01背包，题意很清晰啊。

对于背包问题我有一个建议就是都做题。

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27979    Accepted Submission(s): 11404

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

/*问题分析：第二行输入骨头的value[1001]，第二行输入骨头的volume[1001]我们要得到状态方程dp[i][j]=max(dp[i-1][j],dp[i-1][j-volume[i]]+value[i])方程解释：当装入第i个骨头时我们应该考虑到装入第i-1个骨头的价值dp[i][j],我们要比较装入第i与没有装第i个的价值的比较。没有装入第i个就是将第i-1个装入体积为j的包中当装入第i个时就是把第i-1个装入体积为j-volume[i]的包裹中，但是由于装入了第i个所以它的的价值增加了value[i];*/#include<iostream>#include<cstring>using namespace std;int max(int a,int b){  if(a>b)  return a;  else  return b;}int main(int i,int j){  int t;//测试样例  int n;//总骨头数目  int v;//总的包的体积  int static dp[1001][1001];//最后的最大价值  int value[1001];//单个骨头的价值  int volume[1001];//单个骨头的体积  cin>>t;  while(t--)  {    cin>>n>>v;for(i=1;i<=n;i++)cin>>value[i];for(i=1;i<=n;i++)cin>>volume[i];memset(dp,0,sizeof(dp));for(i=1;i<=n;i++){  for(j=0;j<=v;j++)  {  if(j>=volume[i])    dp[i][j]=max(dp[i-1][j],dp[i-1][j-volume[i]]+value[i]);  else  dp[i][j]=dp[i-1][j];  }}cout<<dp[n][v]<<endl;  }  return 0;}