背包

Problem Description

Many years ago , in Teddy’s hometownthere was a man who was called “Bone Collector”. This man like to collectvaries 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 ofcollecting there are a lot of bones , obviously , different bone has differentvalue 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 twointeger N , V, (N <= 1000 , V <= 1000 )representing the number of bonesand the volume of his bag. And the second line contain N integers representingthe value of each bone. The third line contain N integers representing thevolume of each bone.

Output

One integer per line representingthe 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

 

01背包模板题，主要要写出状态转移方程，和递推差不多，求出当前物品的最大值就是要比较下一件物品是放还是不放，将放进去之后的大值存入数组中，之后从数组中查找，就是使用记忆化数组，这样的复杂度会缩小为O(NV)

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int t,n,v,va[1010],vi[1010],dp[1010][1010];void solve(){memset(dp,0,sizeof(dp));int i,j;for(i=n-1;i>=0;i--){for(j=0;j<=v;j++){if(j<vi[i])dp[i][j]=dp[i+1][j];elsedp[i][j]=max(dp[i+1][j],dp[i+1][j-vi[i]]+va[i]);}}//for(i=n-1;i>=0;i--)//{//for(j=0;j<=v;j++)//{//printf("%3d ",dp[i][j]);//}//printf("\n");// } printf("%d\n",dp[0][v]);}int main(){int i;scanf("%d",&t);while(t--){scanf("%d%d",&n,&v);for(i=0;i<n;i++)scanf("%d",&va[i]);for(i=0;i<n;i++)scanf("%d",&vi[i]);solve();}return 0;}