Bone Collector Time Limit: 1000MS Memory Limit: 65536KB 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 231).
Example Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Example Output
14
Hint hdoj2602 http://blog.csdn.net/u014231159/article/details/25605411?utm_source=tuicool&utm_medium=referral Author 翻译（虽然有翻译，但还是建议大家自己翻译试试）：http://fanyi.baidu.com/?aldtype=23#en/zh/
think: 这又是一道英文题，其实自己不太喜欢做英文题，因为有的地方读不懂的感觉实在是不好受，但我们都需要挑战自己，为将来打下基础。 回归题目，这是一道背包问题，大意就是给出了几组测试例子，然后又给出了骨头的数量和骨收集器的体积，然后接着是两行，第一行包含每个骨头的价值，第二行包含每个骨头的体积，要求计算出骨收集器的最大总价值。 代码实现：
#include<stdio.h>int max(int a, int b){ if(a>b) return a; else return b;}int main(){ int x, n; int p[100005], m[100050]; int sum[200000]; int t; scanf("%d", &t); while(t--) { scanf("%d %d", &n, &x); int i, j; memset(sum, 0, sizeof(sum)); for(i=0;i<n;i++) { scanf("%d", &m[i]); } for(i=0;i<n;i++) { scanf("%d", &p[i]); } for(i=0;i<n;i++) { for(j=x;j>=0;j--) { if(j>=p[i]) sum[j] = max(sum[j-p[i]]+m[i], sum[j]); } } printf("%d/n", sum[x]); } return 0;}新闻热点
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