hdu2602 Bone Collector--01背包

原题链接： http://acm.hdu.edu.cn/showproblem.php?pid=2602

一：原题内容

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

二：分析理解

dp[n][v]表示在容积为v的情况下，放进前n个骨头的最大重量。

三：AC代码

#include<iostream>  #include<algorithm>  using namespace std;int T;int n, v;int va[1005];int vo[1005];int dp[1005][1005];int main(){scanf("%d", &T);while (T--){scanf("%d%d", &n, &v);memset(dp, 0, sizeof(dp));for (int i = 1; i <= n; i++)scanf("%d", &va[i]);for (int i = 1; i <= n; i++)scanf("%d", &vo[i]);for (int i = 1; i <= n; i++){for (int j = 0; j <= v; j++){if (j - vo[i] >= 0)dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - vo[i]] + va[i]);elsedp[i][j] = dp[i - 1][j];}}printf("%d\n", dp[n][v]);}return 0;}

参考博客： http://www.cnblogs.com/Su-Blog/archive/2012/08/28/2659872.html