HDU2016 A

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

15 101 2 3 4 55 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