大家好，又见面了，我是全栈君。
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 ?
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.
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).
31).
Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14
先输入一个t，代表有t个样例。然后输入n和v，n代表有多少骨头。v代表背包体积，每样东西仅仅有一个，也就是说仅仅能取一次，刚開始看这道题的时候全然不知道怎么做。感觉会非常麻烦的样子，学长是作为例题给我们讲的。刚開始有点头绪，但详细还是全然不懂。后来自己专门刷这方面的专题。还算理解的比較好。动态规划的思想得要理解好，这题是用空间换时间。过程中会产生大量之间数据，嗯，就是这样。好了，回到这道题。我们用一维数组来存储数据，可是有些pdf文档比方背包九讲就是用二维数组来解说，都能够的。
如今进入正题：输入前面讲过了，如今讲核心代码
for(i=1;i<=n;i++)
for(j=v;j>=c[i];j--)
liu[j]=max(liu[j],liu[j-c[i]]+w[i]);
这三行就是核心。就像背包九讲里面说的，这个状态转移方程很重要，一定要理解，它联系了上一个状态和这一个状态，所以叫做状态转移方程！
！！！
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为了更加清楚描写叙述执行过程中数组每一个值的详细变化，我在这里加了几行代码：
for(i=1;i<=n;i++)
{
for(j=v;j>=c[i];j--)
{
liu[j]=max(liu[j],liu[j-c[i]]+w[i]);
}
for(k=1;k<=v;k++)
printf("%d ",liu[k]);
printf("\n");
}
我们以题目给的数据为例，执行结果例如以下：
从图中我们能够看出（结合上面的代码），程序循环五次，每次循环的结果都在动态变化。假设还不能理解，建议自己在草稿子上模拟i=1。i=2的时候数组变化的情况，就会非常好理解了。动态规划给我的感觉就是代码非常短，可是理解之后就非常easy了！！！
！！！
好了。讲完了，近期在做dp专题，会不定时更新做题的心得还有题解，大家一起学习。
以下贴下AC代码：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
int main()
{
int i,j,k;
int t,n,m,v;
int liu[1006],c[1006],w[1006];
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&v);
for(i=1;i<=n;i++)
scanf("%d",&w[i]);
for(i=1;i<=n;i++)
scanf("%d",&c[i]);
memset(liu,0,sizeof(liu));
for(i=1;i<=n;i++)
{
for(j=v;j>=c[i];j--)
{
liu[j]=max(liu[j],liu[j-c[i]]+w[i]);
}
for(k=1;k<=v;k++)
printf("%d ",liu[k]);
printf("\n");
}
printf("%d\n",liu[v]);
}
return 0;
}
写代码能力有限，如有编程爱好者发现bug，还请指出，不胜感激！
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