HDU 2602 Bone Collector（01背包）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602点击打开链接

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 66521    Accepted Submission(s): 27756

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

 

基础01背包

#include<bits/stdc++.h>#define maxn 1010using namespace std;struct xjy{    int value;    int volume;    bool operator < (const xjy &r)const    {        return value>r.value;    }};xjy a[maxn];int dp[maxn];int main(){    int n;    int sum=0;    while(cin >> n)    {        while(n--)        {            memset(dp,0,sizeof(dp));            int num;int V;            cin >> num >> V;            for(int i=1;i<=num;i++)                scanf("%d",&a[i].value);            for(int i=1;i<=num;i++)                scanf("%d",&a[i].volume);            sort(a+1,a+num+1);            for(int i=1;i<=num;i++)                for(int j=V;j>=a[i].volume;j--)                    dp[j]=max(dp[j],dp[j-a[i].volume]+a[i].value);            cout << dp[V] << endl;        }    }}