标签：acm dp hdu 01背包
Bone Collector
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
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
裸的01背包啦啦啦啦
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1005;
int n, v, d[N], val[N], vol[N], cas;
int main()
{
scanf ("%d", &cas);
while (cas--)
{
memset (d, 0, sizeof (d));
scanf ("%d%d", &n, &v);
for (int i = 1; i <= n; ++i)
scanf ("%d", &val[i]);
for (int i = 1; i <= n; ++i)
{
scanf ("%d", &vol[i]);
for (int j = v; j >= vol[i]; --j)
d[j] = max (d[j], d[j - vol[i]] + val[i]);
}
printf ("%d\n", d[v]);
}
return 0;
}
HDU 2602 Bone Collector（01背包）,布布扣,bubuko.com
HDU 2602 Bone Collector（01背包）
标签：acm dp hdu 01背包
原文地址：http://blog.csdn.net/iooden/article/details/38458115
