题目：
Given a set of n items, each with a weight w[i] and a value v[i], determine a way to choose the items into a knapsack so that the total weight is less than or equal to a given limit B and the total value is as large as possible. Find the maximum total value. (Note that each item can be only chosen once).
Input
The first line contains the integer T indicating to the number of test cases.
For each test case, the first line contains the integers n and B.
Following n lines provide the information of each item.
The i-th line contains the weight w[i] and the value v[i] of the i-th item respectively.
1 <= number of test cases <= 100
1 <= n <= 500
1 <= B, w[i] <= 1000000000
1 <= v[1]+v[2]+...+v[n] <= 5000
All the inputs are integers.
Output
For each test case, output the maximum value.
Sample Input
1 5 15 12 4 2 2 1 1 4 10 1 2
Sample Output
15
解题思路：就是很裸的01背包问题，然而 他给的体积的数据范围确实太大了，根本没办法去开大数组求解，所以 change一下思维 ，我们将价值作为下标，这样就不会因为开过大数组而爆内存，同时需要注意的一点就是，之前以体积作为下标的时候，我们是求解的固定体积能达到的最大价值，现在就是当前价值能实现的最小体积。当当前的体积是小于给出固定体积的时候，咱们就记录一下，最后直接倒着推，就能找到当前体积的最大价值了。
ac代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int min(int a,int b)
{
if(a<b)
return a;
else
return b;
}
typedef long long ll;
ll w[505];
int v[505];
int dp[80008];
int main()
{
int t;
while(scanf("%d",&t)!=EOF)
{
int n,B;
while(t--)
{
int vv=0;
memset(dp,0x3f3f,sizeof(dp));
scanf("%d%d",&n,&B);
for(int i=0;i<n;i++)
scanf("%lld%d",&w[i],&v[i]);
for(int i=0;i<n;i++)
vv+=v[i];
dp[0]=0;
for(int i=0;i<n;i++)
for(int j=vv;j>=v[i];j--)
{
if(dp[j-v[i]]+w[i]<=B)
{
dp[j]=min(dp[j-v[i]]+w[i],dp[j]);
}
}
int ans=-1;
for(int j=vv;j>=0;j--)
{
if(dp[j]<=B)
{
ans=j;
break;
}
}
cout<<ans<<endl;
}
}
return 0;
}总结：不得不说，这个真的是开了一个转化思维的脑洞，之前习惯将体积做下标，价值做数值，颠倒过来其实也是很不错的操作。