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 2
31).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
Author
Teddy
基础01背包
#include<bits/stdc++.h>
#define maxn 1010
using 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;
}
}
}
