hdu 2602 01背包

Bone Collector Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 56935    Accepted Submission(s): 23782 [align=left]Problem Description[/align] 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 ?   [align=left]Input[/align] 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.   [align=left]Output[/align] One integer per line representing the maximum of the total value (this number will be less than 231).   [align=left]Sample Input[/align] 15 101 2 3 4 55 4 3 2 1   [align=left]Sample Output[/align] 14

动态规划先找出子问题，我们可以这样考虑：在物品比较少，背包容量比较小时怎么解决？用一个数组f[i][j]表示，在只有i个物品，容量为j的情况下背包问题的最优解，那么当物品种类变大为i+1时，最优解是什么？第i+1个物品可以选择放进背包或者不放进背包，假设放进背包（前提是放得下），那么f[i+1][j]=f[i][j-weight[i+1]+value[i+1]；如果不放进背包，那么f[i+1][j]=f[i][j]。放不下就是f[i+1][j]=f[i][j];

这就得出了状态转移方程：

f[i+1][j]=max(f[i][j],f[i][j-weight[i+1]+value[i+1])

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int f[1010][1010],weight[1010],value[1010];
int main()
{
int t,n,m,i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&n,&m);
for(i=1;i<=n;i++)
scanf("%d",&value[i]);
for(i=1;i<=n;i++)
scanf("%d",&weight[i]);
for(i=1;i<=n;i++)
{
for(j=0;j<=m;j++)//骨头的占用空间可能为0；
{
if(weight[i]<=j)
f[i][j]=max(f[i-1][j],f[i-1][j-weight[i]]+value[i]);
else f[i][j]=f[i-1][j];
}
}
printf("%d\n",f
[m]);
}
return 0;
}

上面计算f[i][j]可以看出，在计算f[i][j]时只使用了f[i-1][0……j]，没有使用其他子问题，因此在存储子问题的解时，只存储f[i-1]子问题的解即可。这样可以用两个一维数组解决，一个存储子问题，一个存储正在解决的子问题。再进一步思考，计算f[i][j]时只使用了f[i-1][0……j]，没有使用f[i-1][j+1]这样的话，我们先计算j的循环时，让j=M……1，只使用一个一维数组即可。这就得出了状态转移方程：
f[j]=max(f[j],f[j-weight[i]+value[i]) ；

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int f[1010],weight[1010],value[1010];//全局变量，自动清0；
int main()
{
int t,n,m,i,j;
scanf("%d",&t);
while(t--)
{
memset(f,0,sizeof(f));//记得清0；
scanf("%d %d",&n,&m);
for(i=1;i<=n;i++)
scanf("%d",&value[i]);
for(i=1;i<=n;i++)
scanf("%d",&weight[i]);
for(i=1;i<=n;i++)
{
for(j=m;j>=weight[i];j--)
{
f[j]=max(f[j],f[j-weight[i]]+value[i]);
}
}
printf("%d\n",f[m]);
}
return 0;
}