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题目：Bone Collector - 时间限制：2000/1000 MS (Java/其他) 内存限制：32768/32768 K (Java/其他) 总提交次数：28365 通过提交次数：11562

题目描述：很多年前，在Teddy的家乡，有一个叫做Bone Collector的游戏。这个游戏的目标是在限定时间内收集尽可能多的骨头。

输入：输入的第一行包含两个整数，分别表示游戏的总时间和每个时间单位内可以收集的骨头数量。

输出：输出一个整数，表示在限定时间内可以收集到的最大骨头数量。

HDU2602BoneCollector【0

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 28365Accepted Submission(s): 11562 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

15 101 2 3 4 55 4 3 2 1

Sample Output

14

01背包入门题。

#include #include #define maxn 1002int dp[maxn], w[maxn], v[maxn];int main(){ int t, n, val, i, j; scanf("%d", while(t--){ scanf("%d%d", for(i = 1; i <= n; ++i) scanf("%d", v + i); for(i = 1; i <= n; ++i) scanf("%d", w + i); memset(dp, 0, sizeof(dp)); for(i = 1; i <= n; ++i){ for(j = val; j >= w[i]; --j){ if(dp[j]