HDU 2062 Bone Collector (01背包)

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 59212 Accepted Submission(s): 24709

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

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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题目大意：背包装有体积和价值的骨头，怎样才能使价值最大？

典型的01背包问题：每个骨头只有一件，可以选择放与不放。背包容量V,骨头体积vol[i]和价值val[i],获得的最大价值dp[v].并没有要求必须把背包

装满，而是只希望价格尽可能大，所以初始化时dp[0...V]全部设为0。

状态转移方程：dp[j] = max(dp[j-vol[i]]+val[i],dp[j]);

#include<stdio.h>#include <algorithm> #include<cstring>using namespace std;int main(){int T;   //输入数据组数 scanf("%d",&T);while(T--){int N, V; //定义数量  和总体积scanf("%d%d",&N,&V);int i,j,val[1005],vol[1005],dp[1005];for(i = 0; i < N; i++){scanf("%d",&val[i]); } for( i = 0; i < N; i++){scanf("%d",&vol[i]);}memset(dp,0,sizeof(dp));  //初始化 for(i = 0; i < N; i++){for(j = V; j >= vol[i];j--){dp[j] = max(dp[j-vol[i]]+val[i],dp[j]);}}printf("%d\n",dp[V]);} return 0;}//一定注意输出数据个数   #include<algorithm>

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