HDU 2602 dp

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5029    Accepted Submission(s): 1764

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

/* 典型的01揹包問題,c[i]爲第i件物品的價值，w[i]爲第i件物品的體積，dp[j]表示容量爲j時能獲得的最大價值 */ #include<stdio.h> int dp[1005]; int c[1005],w[1005]; int main() { int t; int i,j; int n,v; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&v); for(i=1;i<=n;i++) { scanf("%d",&c[i]); } for(i=1;i<=n;i++) { scanf("%d",&w[i]); } for(i=0;i<=v;i++) // 這裏是i<=v dp[i]=0; for(i=1;i<=n;i++) for(j=v;j>=w[i];j--) if(dp[j]<dp[j-w[i]]+c[i]) dp[j]=dp[j-w[i]]+c[i]; printf("%d/n",dp[v]); } return 0; }