C - Bone Collector(背包问题)（01背包）

C - Bone Collector

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 2 ³¹).

Sample Input

Sample Output

嗯一道简单的01背包问题

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define maxn 1005
int over[maxn];
int volume[maxn];
int value[maxn];
int main()
{int t;scanf("%d",&t);while(t--){int n,m,i,j;scanf("%d%d",&n,&m);memset(value,0,sizeof(value));memset(volume,0,sizeof(volume));memset(over,0,sizeof(over));for( i=1;i<=n;i++)scanf("%d",&value[i]);for( i=1;i<=n;i++)scanf("%d",&volume[i]);for(i=1;i<=n;i++){for(j=m;j>=volume[i];j--){over[j]=max(over[j],over[j-volume[i]]+value[i]);}}printf("%d\n",over[m]);}
return 0;
}

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C - Bone Collector(背包问题)（01背包）

C - Bone Collector

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