题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=2602
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 77450 Accepted Submission(s): 32095
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
分析：
经典的01背包问题，每个物品只有两种状态，放还是不放
dp[i][j]:即前面i件物品放入一个容量为j的背包可以获得的最大价值
状态转移方程：
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] j>=w[i]
注意：
1.先输入的是物品的价值，而不是重量，题目中好像说反了
2.背包容量可以为0，物品的重量也可以为0
关于第2点的测试数据：
输入
1
5 0
2 4 1 5 1
0 0 1 0 0
2 4 1 5 1
0 0 1 0 0
输出
12
从前往后遍历二维数组
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,c;
scanf("%d %d",&n,&c);
int v[n+],w[n+];
for(int i=;i<=n;i++)
{
scanf("%d",&v[i]);
}
for(int i=;i<=n;i++)
{
scanf("%d",&w[i]);
}
int dp[n+][c+];
memset(dp,,sizeof(dp));
for(int i=;i<=n;i++)
{
for(int j=;j<=c;j++)
{
if(w[i]<=j)//表示第i个物品放入背包中
{
dp[i][j]=max(dp[i-][j],dp[i-][j-w[i]]+v[i]);//第i个物品放入之后，那么前面i-1个物品可能会因为剩余空间不够无法放入
}else//表示第i个物品不放入背包
{
dp[i][j]=dp[i-][j];//如果第i个物品不放入背包，那么此时的最大价值与放前面i-1个物品的值相等
}
}
}
printf("%d\n",dp[n][c]);
}
return ;
} /*
3
5 10
6 3 5 4 6
2 2 6 5 4
5 10
1 2 3 4 5
5 4 3 2 1
5 0
2 4 1 5 1
0 0 1 0 0 15
14
12
*/
从后往前遍历二维数组
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,c;
scanf("%d %d",&n,&c);
int v[n+],w[n+];
for(int i=;i<=n;i++)
{
scanf("%d",&v[i]);
}
for(int i=;i<=n;i++)
{
scanf("%d",&w[i]);
}
int dp[n+][c+];
memset(dp,,sizeof(dp));
for(int j=;j<=c;j++)
{
if(j>=w[n])
{
dp[n][j]=v[n];
}else
{
dp[n][j]=;
}
}
for(int i=n-;i>=;i--)
{
for(int j=;j<=c;j++)
{
if(j>=w[i])
{
dp[i][j]=max(dp[i+][j],dp[i+][j-w[i]]+v[i]);
}
else
{
dp[i][j]=dp[i+][j];
}
}
}
printf("%d\n",dp[][c]);
}
return ;
} /*
3
5 10
6 3 5 4 6
2 2 6 5 4
5 10
1 2 3 4 5
5 4 3 2 1
5 0
2 4 1 5 1
0 0 1 0 0 15
14
12
*/
二者其实是一个方法，只是遍历二维数组的方向有点不同