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
坑点在于有一种物体体积为0==
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
struct node
{
int len, large;
};
int dp[1000][1005];
node goods[1005];
int main()
{
int n;
scanf("%d", &n);
while (n--)
{
memset(dp, 0, sizeof(dp));
memset(goods, 0, sizeof(goods));
int num, large;
scanf("%d %d", &num, &large);
for (int i = 0; i < num; i++)
{
scanf("%d", &goods[i].large);
}
for (int i = 0; i < num; i++)
{
scanf("%d", &goods[i].len);
}
for (int i = 0; i < num; i++)
{
for (int j = 0; j <=large; j++)
{
if (i == 0)
{
if (j >= goods[i].len )
{
dp[i][j] = goods[i].large;
}
else
{
dp[i][j] = 0;
}
}
else
{
if (j >= goods[i].len)
{
dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - goods[i].len] + goods[i].large);
}
else
{
dp[i][j] = dp[i - 1][j];
}
}
}
}
printf("%d\n", dp[num - 1][large]);
}
return 0;
}//by swust t_p
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