Bone Collector-优快云博客

本文链接：https://blog.youkuaiyun.com/zhang_di233/article/details/43821167

本文深入探讨了骨收集者背包问题，一个典型的01背包问题实例。通过具体输入输出示例，详细解释了解决该问题的01背包算法，并提供了关键代码实现，帮助读者理解如何解决此类优化问题。

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A - Bone Collector

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status

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 2 ³¹).

Sample Input

Sample Output

这个题就属于典型的01背包问题。

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
int dp[10000];
int bonev[10000];         //记录每个物品的价值
int bonew[10000];         //记录每个物品的体积
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,v,i,j;
        memset(dp,0,sizeof(dp));
        scanf("%d %d",&n,&v);
        for(i=1; i<=n; i++)
            scanf("%d",&bonev[i]);
        for(i=1; i<=n; i++)
            scanf("%d",&bonew[i]);
        for(i=1; i<=n; i++)       //遍历每个物品
            for(j=v; j>=bonew[i]; j--)       //对于01背包问题，体积是从大到小遍历，而完全背包则是从小到大遍历 
                dp[j]=max(dp[j-bonew[i]]+bonev[i],dp[j]);
        printf("%d\n",dp[v]);

    }
    return 0;
}