本文主要是介绍Bone Collector -01背包，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目描述
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 ?
输入
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.
输出
One integer per line representing the maximum of the total value (this number will be less than 231).
示例输入
1 5 10 1 2 3 4 5 5 4 3 2 1
示例输出
14
提示
hdoj2602
来源
#include<stdio.h>
#include<string.h>int dp[1010][1010];
int p[1010],w[1010];int qmax (int a,int b)
{return a > b ? a : b;
}int main()
{int N;int i,k,l,v;scanf ("%d",&N);while (N--){memset (dp,0,sizeof (dp));scanf ("%d%d",&l,&v);for (i = 1; i <= l; i++)scanf ("%d",&w[i]);for (i = 1; i <= l; i++)scanf ("%d",&p[i]);for (i = 1; i <= l; i++)for (k = 0; k <= v; k++){if (k >= p[i])dp[i][k] = qmax (dp[i - 1][k],dp[i - 1][k - p[i]] + w[i]);elsedp[i][k] = dp[i - 1][k];}printf ("%d\n",dp[l][v]);}return 0;
}
这篇关于Bone Collector -01背包的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
