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HDU 2602 Bone Collector
http://acm.hdu.edu.cn/showproblem.php?pid=2602
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 32408 Accepted Submission(s): 13329
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
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14
分析：
http://blog.csdn.net/dapengbusi/article/details/7463968
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 int v[1000],w[1005]; 6 int n,c; 7 int f[1005][1005];//对于dp(i,j)就表示可选物品为i…n背包容量为j(总重量)时背包中所放物品的最大价值。 8 void backpack() 9 {10 int i,j;11 for(i=1;i<=n;i++)12 for(j=0;j<=c;j++)13 {14 if(i==1)15 f[i][j]=0;16 else17 f[i][j]=f[i-1][j];18 if(j>=v[i])19 f[i][j]=max(f[i][j],f[i-1][j-v[i]]+w[i]);20 }21 printf("%d\n",f[n][c]);22 }23 int main()24 {25 int t;26 scanf("%d",&t);27 while(t--)28 {29 int i;30 scanf("%d%d",&n,&c);31 for(i=1;i<=n;i++)32 scanf("%d",&w[i]);33 for(i=1;i<=n;i++)34 scanf("%d",&v[i]);35 backpack();36 }37 return 0;38 }
HDU 2602 Bone Collector
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