hdu 2602 Bone Collector（dp46）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 53210    Accepted Submission(s): 22383

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).

 

Sample Input

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14
题目大意：有一个收集骨头的人，有一个V容量的包裹，他想要尽可能收集总价值最大的骨头，问你最多能收藏价值多少的骨头。
题目分析：这是一个比较典型的01背包问题，需要注意的是数要用long long型。然后，，，好像就没有然后了。贴代码，哦对了，那个状态转移方程是dp【j】=max{dp【j-a【i】】+b【i】，dp【j】}；嗯，就是这样。
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>using namespace std;#define maxn 1005long long  dp[maxn],a[maxn],b[maxn];int main(){int n,m,t;while(~(scanf("%d",&t))){while(t--){memset(dp,0,sizeof(dp));scanf("%d%d",&n,&m);for(int i=0;i<n;i++)   scanf("%I64d",&b[i]);for(int i=0;i<n;i++)   scanf("%I64d",&a[i]);for(int i=0;i<n;i++)   for(int j=m;j>=a[i];j--){   dp[j]=max(dp[j-a[i]]+b[i],dp[j]);   }printf("%lld\n",dp[m]);}}return 0;}