hdu2602 (01背包)

最新推荐文章于 2024-03-28 20:10:23 发布

hansy2015

最新推荐文章于 2024-03-28 20:10:23 发布

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本文链接：https://blog.youkuaiyun.com/gh6267/article/details/77169248

本文介绍了一种典型的背包问题——骨收集者问题。该问题是关于如何最大化背包内物品总价值的经典案例，通过输入不同骨头的价值与体积，利用动态规划算法求解最大总价值。

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 65765 Accepted Submission(s): 27411

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
1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output
14

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;

int v[1111],w[1111],dp[1111];

void beg01(int V,int v,int w){
    for(int i=V;i>=v;i--){
        dp[i]=max(dp[i],dp[i-v]+w);
    }
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        memset(dp,0,sizeof(dp));
        int N,V;
        scanf("%d %d",&N,&V);
        for(int i=1;i<=N;i++){
            scanf("%d",&w[i]);
        }
        for(int i=1;i<=N;i++){
            scanf("%d",&v[i]);
        }
        for(int i=1;i<=N;i++){
            beg01(V,v[i],w[i]);
        }
        printf("%d\n",dp[V]);
    }
    return 0;
}