HDU 2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 59344    Accepted Submission(s): 24773

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

解题思路：题目大意就是一个人他要往包里面装东西，使得价值最大。先输入价值，在输入体积。。。。这就是一个简单的01背包

代码如下：

#include <stdio.h>#include <math.h>#include <string.h>#include <algorithm>using namespace std;struct P{    int x,y;///分别表示价值和体积}a[1010];int main(){    int n,v,t,dp[2020];    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        scanf("%d %d",&n,&v);        for(int i=0;i<n;i++)            scanf("%d",&a[i].x);        for(int i=0;i<n;i++)            scanf("%d",&a[i].y);        for(int i=0;i<n;i++)            for(int j=v;j>=a[i].y;j--)                dp[j]=max(dp[j],dp[j - a[i].y]+a[i].x);        printf("%d\n",dp[v]);    }    return 0;}