杭电2602（01背包问题）——第一次做背包问题的理解
Bone Collector
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 231).
Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14
分析：题的大意就是一个家伙喜欢收集骨头，让你求一下在容器有限的情况下，装的骨头的最大价值；
例如给的输入数据是：
1//表示测试数据有一组
5 10//5表示有有五个骨头，10表示容器容量为10
1 2 3 4 5//表示五个骨头每个的价值
5 4 3 2 1//表示五个骨头的体积
5 10//5表示有有五个骨头，10表示容器容量为10
1 2 3 4 5//表示五个骨头每个的价值
5 4 3 2 1//表示五个骨头的体积
输出为
14//装价值为5,4,3,2,的骨头价值最大。
01背包算法：
01背包的表：
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 0 | 0 | 0 | 2 | 2 | 2 | 2 | 2 | 3 | 3 |
| 3 | 0 | 0 | 3 | 3 | 3 | 3 | 5 | 5 | 5 | 5 |
| 4 | 0 | 4 | 4 | 4 | 7 | 7 | 7 | 7 | 9 | 9 |
| 5 | 5 | 5 | 9 | 9 | 9 | 12 | 12 | 12 | 12 | 14 |
上表的我理解的意思就是：(第一行是j的值变化，第一列是i的值变化）
骨头编号按价值从1—5编号i=1,2 3,4,5. j时容量从1到10变化
i=0时全是0；
当i=1时，j从0到10变化，看能装的最大价值，可见当容器为5时，能装下此时价值i=1，体积为5的骨头，
当j=6,7,8,9,10时只能装上5，要问为6时为什么不要装容价值为1和容量为1的i=5的骨头呢？是因为此时
还没有排到她，此表是从容量为1，1号骨头开始往后排的，是此时此空间内所能放骨头的最大值。
然后在逐渐加大i和j的值，再在所增加的空间上最大的价值，怎样保证最大呢？就是在所增加的空间上先放一个最大价值的骨头，在求出剩余的空间，从前面空间内找到剩余空间上放的价值，加到一起就是最大的价值（因为剩余空间上放的也是最大价值！所以在右下角的14就是最大价值！当然也有一些情况是如果加上最大价值的那个所得价值不是最大的，那就比较一下，是不是最大的，怎样比较呢？就是记下每一次增加的最大值，与再增加的比较就行了！
代码如下：
#include<iostream>
using namespace std;
int a[1005],b[1005];
int d[1005][1005];
int main()
{
int n;
cin>>n;
while(n--)
{ int x,y,i,j;
cin>>x>>y;
for(i=1;i<=x;i++)
cin>>a[i];
for(i=1;i<=x;i++)
cin>>b[i];
for(i=1;i<=x;i++)
for(j=0;j<=y;j++)
{
if(j>=b[i])
{
d[i][j]=d[i-1][j]>(d[i-1][j-b[i]]+a[i])?d[i-1][j]:(d[i-1][j-b[i]]+a[i]);//核心代码，也是公式
}
else
d[i][j]=d[i-1][j];
}
cout<<d[x][y]<<endl;
}
return 0;
}
using namespace std;
int a[1005],b[1005];
int d[1005][1005];
int main()
{
int n;
cin>>n;
while(n--)
{ int x,y,i,j;
cin>>x>>y;
for(i=1;i<=x;i++)
cin>>a[i];
for(i=1;i<=x;i++)
cin>>b[i];
for(i=1;i<=x;i++)
for(j=0;j<=y;j++)
{
if(j>=b[i])
{
d[i][j]=d[i-1][j]>(d[i-1][j-b[i]]+a[i])?d[i-1][j]:(d[i-1][j-b[i]]+a[i]);//核心代码，也是公式
}
else
d[i][j]=d[i-1][j];
}
cout<<d[x][y]<<endl;
}
return 0;
}