Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51860 Accepted Submission(s): 21837
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
题目地址：
点击打开链接
放松一下：
陈奕迅《你的背包》
AC代码：
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
int n, v, val[1005];
int cost[1005], f[1005];
int main()
{
int t;scanf("%d", &t);
while(t--)
{
int i, j;
scanf("%d %d", &n, &v);
memset(f, 0, sizeof(f));
for(i = 1; i <= n; i++)
scanf("%d", &val[i]);
for(i = 1; i <= n; i++)
scanf("%d", &cost[i]);
for(i = 1; i <= n; i++)
{
for(j = v; j >= cost[i]; j--)
{
f[j] = max(f[j], f[j-cost[i]]+val[i]);
}
}
printf("%d\n", f[v]);
}
return 0;
}
Sample Output
14
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