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Problem DescriptionGiven 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).
InputThe 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.
OutputFor each test case, output the maximum value.
Sample Input Sample Output/*
0-1背包问题：
因为重量太大，肯定不能按照原来的标准模型套用，开不了那么大的数组
但是总价值却小于5000;
根据价值构造dp方程,找到最小的重量能够得到这个价值
dp[100]=15,就是代表使得价值为100的最小重量为15;
*/
#include <iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxs = 600;
int n,B;
int w[maxs];
int v[maxs];
int dp[5005];
int main()
{
freopen("in.txt","r",stdin);
int T;
cin>>T;
while(T--)
{
int sum = 0;
scanf("%d%d\n",&n,&B);
for(int i=1;i<=n;i++)
scanf("%d%d",&w[i],&v[i]),sum+=v[i];
memset(dp,0x3f,sizeof(dp));
dp[0]=0;
for(int i=n;i>=1;i--)
for(int j=sum;j>=v[i];j--)
dp[j]=min(dp[j],dp[j-v[i]]+w[i]);
for(int i=sum;i>=0;i--)
if(dp[i]<=B)
{
printf("%d\n",i);
break;
}
}
return 0;
}
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原文地址：http://www.cnblogs.com/wt20/p/5794355.html
