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
Author
Teddy
Source
题意：给你n个骨头的价值和体积，求体积为V时最多有多少价值；
01背包问题:
递推公式：dp[i][j]=dp[i+1][j];(j<w[i]);
dp[i][j]=max(dp[i+1][j],dp[i+1][j-w[i]]+v[i]);
代码如下：
关于i的逆向循环；
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define N 1100
int dp[N][N];
int main()
{
int T,i,j,V,n,v[N],w[N]; scanf("%d",&T);
while(T--)
{
memset(dp,,sizeof(dp));
scanf("%d%d",&n,&V);
for(i=;i<n;i++)
scanf("%d",&v[i]);
for(i=;i<n;i++)
scanf("%d",&w[i]);
for(i=n-;i>=;i--)
{
for(j=;j<=V;j++)
{
if(j<w[i])
dp[i][j]=dp[i+][j];
else
dp[i][j]=max(dp[i+][j],dp[i+][j-w[i]]+v[i]);
}
}
printf("%d\n",dp[][V]);
}
return ;
}
下面是关于i的正向循环；
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define N 1100
int dp[N][N];
int main()
{
int T,i,j,V,n,v[N],w[N]; scanf("%d",&T);
while(T--)
{
memset(dp,,sizeof(dp));
scanf("%d%d",&n,&V);
for(i=;i<n;i++)
scanf("%d",&v[i]);
for(i=;i<n;i++)
scanf("%d",&w[i]);
for(i=;i<n;i++)
{
for(j=;j<=V;j++)
{
if(j<w[i])
dp[i+][j]=dp[i][j];
else
dp[i+][j]=max(dp[i][j],dp[i][j-w[i]]+v[i]);
}
}
printf("%d\n",dp[n][V]);
}
return ;
}
用一维数组；
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define N 1100
int dp[N];
int main()
{
int T,i,j,V,n,v[N],w[N]; scanf("%d",&T);
while(T--)
{
memset(dp,,sizeof(dp));
scanf("%d%d",&n,&V);
for(i=;i<n;i++)
scanf("%d",&v[i]);
for(i=;i<n;i++)
scanf("%d",&w[i]);
for(i=;i<n;i++)
{
for(j=V;j>=w[i];j--)
{
dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
}
}
printf("%d\n",dp[V]);
}
return ;
}