HDU--2602 （0-1背包）
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
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
#include<iostream>
using namespace std;
#define N 1010
int w[N],v[N],f[N][N]; //f[i][j]---背包容量为j存放前i件物品的最大价值
int main()
{
int vol,i,num,t; //num---物品的数量,vol---背包的容量,value--背包的最大价值
cin>>t;
while(t--)
{
memset(f,0,sizeof(f));
cin>>num>>vol;
for(i=1;i<=num;i++)
cin>>v[i];
for(i=1;i<=num;i++)
cin>>w[i];
for(i=1;i<=num;i++)
for(int j=vol;j>=0;j--)
{
if(j>=w[i]) f[i][j]=max(f[i-1][j],f[i-1][j-w[i]]+v[i]);
else f[i][j]=f[i-1][j];
}
cout<<f[num][vol]<<endl;
}
return 0;
}
注：全局变量在静态存储区分配内存，局部变量是在栈上分配内存空间的
VC堆栈默认是1M
上述代码也可用一维数组（备忘录记录区--dp[N]）
#include<iostream>
using namespace std;
#define N 1010
int w[N],v[N],dp[N]; //f[i][j]---背包容量为j存放前i件物品的最大价值
int main()
{
int vol,i,num,t; //num---物品的数量,vol---背包的容量,value--背包的最大价值
cin>>t;
while(t--)
{
memset(dp,0,sizeof(dp));
cin>>num>>vol;
for(i=1;i<=num;i++)
cin>>v[i];
for(i=1;i<=num;i++)
cin>>w[i];
for(i=1;i<=num;i++)
for(int j=vol;j>=w[i];j--)
{
dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
}
cout<<dp[vol]<<endl;
}
return 0;
}