【题解】HDU-2602 Bone Collector

Bone Collector (HDU - 2602)

题面

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 ?

输入

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.

输出

One integer per line representing the maximum of the total value (this number will be less than 2^31).

样例输入

1
2
3
4
1
5 10
1 2 3 4 5
5 4 3 2 1

样例输出

1
14

提示

思路

代码

查看代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 1e3+5;

int T, n, m;
int c[N], w[N];
int dp[N];

int main(void) {
scanf("%d", &T);
while(T--) {
scanf("%d%d", &n, &m);
for(int i=0; i<n; i++) {
scanf("%d", &w[i]);
}
for(int i=0; i<n; i++) {
scanf("%d", &c[i]);
}
memset(dp, 0, sizeof(dp));
for(int i=0; i<n; i++) {
for(int j=m; j>=c[i]; j--) {
dp[j] = max(dp[j], dp[j-c[i]]+w[i]);
}
}
printf("%d\n", dp[m]);
}
return 0;
}
_/_/_/_/_/ EOF _/_/_/_/_/