HDU 2602 Bone Collector(01背包裸题)

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

思路：

动规01 背包模板题，真的是裸的不能再裸了。

代码：

#include<stdio.h>#include<iostream>#include<queue>#include<cstring>#include<cmath>using namespace std;#define MAXN 1005int N,V;int vol[MAXN],val[MAXN];int dp[MAXN];int main(){int T;scanf("%d",&T);while(T--){memset(dp,0,sizeof(dp));scanf("%d %d",&N,&V);for(int i=0 ; i<N ; i++){scanf("%d",&val[i]);}for(int i=0 ; i<N ; i++){scanf("%d",&vol[i]);}for(int i=0 ; i<N ; i++){for(int j=V ; j>=vol[i] ; j--){dp[j] = max(dp[j],dp[j-vol[i]]+val[i]);}}printf("%d\n",dp[V]);}return 0;}