HDU 2602 Bone Collector 最简单背包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 65995    Accepted Submission(s): 27523

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

这个真的是一点弯弯都没有，所以就直接代码走起来.

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define N 1005using namespace std;int n,m;int dp[N],w[N],v[N];int main(){int ccase,i,j;scanf("%d",&ccase);while(ccase--){memset(dp,0,sizeof(dp));memset(w,0,sizeof(w));memset(v,0,sizeof(v));scanf("%d %d",&n,&m);for(i=1;i<=n;i++) scanf("%d",&v[i]);for(i=1;i<=n;i++) scanf("%d",&w[i]);for(i=1;i<=n;i++){for(j=m;j>=w[i];j--){dp[j] = dp[j] > dp[j - w[i]] + v[i] ? dp[j] : dp[j - w[i]] + v[i];//dp[j]=max(dp[j],dp[j-w[i]]+v[i]);}}printf("%d\n",dp[m]);}}