Bone Collector -01背包

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

   

15 101 2 3 4 55 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]);                else                    dp[i][k] = dp[i - 1][k];            }        printf ("%d\n",dp[l][v]);    }    return 0;}