Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 51847 Accepted Submission(s): 21829
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
Author
Teddy
Source
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状态转移方程：dp[i][j] = max{dp[i+1][j], dp[i+1][j-vol[i]]+val[i]};其中i表示第i个物品，j表示背包当前的容量，而数组中所存放的元素为在当前状态下的最大价值。通过下面的每个状态的最大值向上推。因此，最终的最大状态应该存放在dp[0][v]中。
#include#include #include #include using namespace std;const int MAXN = 1000 + 10;const int INF = 0x3f3f3f3f;int n, v, val[MAXN], vol[MAXN], dp[MAXN][MAXN];int main() { int t; scanf("%d", &t); while (t--) { scanf("%d %d", &n, &v); for (int i = 0; i < n; i++) { scanf("%d", &val[i]); } for (int i = 0; i < n; i++) { scanf("%d", &vol[i]); } memset(dp, 0, sizeof(dp)); for (int i = n - 1; i >= 0; i--) { for (int j = 0; j <= v; j++) { if (j < vol[i]) dp[i][j] = dp[i + 1][j]; else dp[i][j] = max(dp[i+1][j], dp[i+1][j-vol[i]]+val[i]); } } printf("%d\n", dp[0][v]); } return 0;}