Bone CollectorTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 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
Sample Output
翻译：有一人爱收集骨头，每个骨头有一个价值和重量，他有一个背包，容量为V，问他装得最大价值量是多少？（经典的01背包问题）
重点：01背包、动态规划
难点：动态方程的书写
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
int T,n,v,i,j,dp[1100],a[1100],b[1100];//a[1100]是价值，b[1100]是重量；
cin>>T;
while(T--)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(dp,0,sizeof(dp));
cin>>n>>v;//n是骨头个数，V是背包容量
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
for(i=1;i<=n;i++)
scanf("%d",&b[i]);
for(i=1;i<=n;i++)
for(j=v;j>=b[i];j--)
{
if(dp[j]<dp[j-b[i]]+a[i])
dp[j]=dp[j-b[i]]+a[i];
}
printf("%d\n",dp[v]);
}
return 0;
}
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hduoj2602Bone Collector,布布扣,bubuko.com
原文地址：http://blog.csdn.net/hpuhjl/article/details/38332587
