HDU2602_Bone Collector(背包/01背包)

时间：2014-08-01 23:07:32 阅读：284 评论：0 收藏：0 [点我收藏+]

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解题报告

题目传送门

题意：

容量为v的大小，物品数n，每个物品有价值和容量，求能装进包的最大价值。

思路：

基础01背包。

dp[j]=max(dp[j],dp[j-c[i]]+w[i])

#include <iostream>
#include <cstring>
#include <cstdio>
#define inf 99999999
using namespace std;

int main()
{
    int t,i,j,n,v,w[1010],c[1010],dp[1010];
    scanf("%d",&t);
    while(t--) {
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&v);
        for(i=1; i<=n; i++)
            scanf("%d",&w[i]);
        for(i=1; i<=n; i++)
            scanf("%d",&c[i]);
        for(i=1; i<=n; i++) {
            for(j=v; j>=c[i]; j--) {
                dp[j]=max(dp[j],dp[j-c[i]]+w[i]);
            }
        }
        printf("%d\n",dp[v]);
    }
    return 0;
}

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29249 Accepted Submission(s): 11967

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 ?
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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