Bone Collector——01背包

 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

 15 101 2 3 4 55 4 3 2 1 

 

Sample Output

 14 

题意：

一个体积为V的背包，用来装体积为volume，价值为value的骨头，问如何装能够让背包里的价值最大。

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>#include <iostream>using namespace std;int N,V;int value[1009];int volume[1009];int f[1001][1001];int fmax(int i,int j){    if(i > j)        return i;    else        return j;}int main(){    int t,i,j,k;    scanf("%d",&t);    while(t--)    {        memset(f,0,sizeof(f));        scanf("%d%d",&N,&V);        for(i = 1; i <= N; i++)        {            scanf("%d",&value[i]);        }        for(i = 1; i <= N; i++)        {            scanf("%d",&volume[i]);        }        for(i = 1;i <= N;i++)        {            for(j = 0;j <= V;j++)            {                if(j >= volume[i])                {                    f[i][j] = fmax(f[i-1][j],f[i-1][j-volume[i]]+value[i]);                }                else                    f[i][j] = f[i-1][j];            }        }        printf("%d\n",f[N][V]);    }    return 0;}