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
思路：
状态转移方程：val[j]=max{val[j],val[j-weight[i]]+p[i]}
val[j]表示：前i个物体在容量为vol最大的装载量
源代码：
1 #include<iostream>
2 #include<algorithm>
3 #include<string>
4 #include<cstring>
5 #include<cmath>
6 using namespace std;
7 #define maxn 1000+10
8 int val[maxn], weight[maxn], p[maxn];
9 int main()
10 {
11 int T;
12 cin >> T;
13 while (T--)
14 {
15 int n, vol;
16 cin >> n >> vol;
17 for (int i = 1; i <= n; i++)
18 {
19 cin >> p[i];
20 }
21 for (int j = 1; j <= n; j++)
22 {
23 cin >> weight[j];
24 }
25 for (int j = 0; j <= vol; j++)
26 {
27 val[j] = 0;
28 }
29 for (int i = 1; i <= n; i++)
30 {
31 for (int j = vol; j >= weight[i]; j--)
32 {
33 val[j] = max(val[j], val[j - weight[i]] + p[i]);
34
35 }
36 }
37 cout << val[vol] << endl;
38
39 }
40 return 0;
41
42 }
心得：
背包入门~~~~理解一个方程还是挺费劲的，/(ㄒoㄒ)/~~不过终于比之前好多了。。。。给自己一个赞！不过还是觉得好神奇
(*^__^*) 嘻嘻……