hdu 2602 01背包入门

题目：

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 62306    Accepted Submission(s): 25961

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

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  1203 2159 1171 2191 1003 

给出n个物品的价值和费用 容量为V的背包疯狂的装物品使得价值最大

资料：

http://love-oriented.com/pack/P01.html

 不一定装满 dp数组初始化为零就可以了

代码：

#include<bits/stdc++.h>using namespace std;const int maxn=1010;int val[maxn],cost[maxn],dp[maxn];int t,n,v;inline void zeroone(int val,int cost){    for(int i=v;i>=cost;--i){        dp[i]=max(dp[i],dp[i-cost]+val);    }}int main(){    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&v);        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;++i) scanf("%d",&val[i]);        for(int i=1;i<=n;++i) scanf("%d",&cost[i]),zeroone(val[i],cost[i]);        printf("%d\n",dp[v]);//not dp[n]    }    return 0;}