01背包（转换思想）（背包容量超大～）

C - Knapsack problem

Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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Description

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

15 1512 42 21 14 101 2

Sample Output

15

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题意 ： 01背包的典型例题，题意就不BB了，此题背包容量过大，内存会超～～

思路： 根据01背包的思想，满足在背包容量以内，用最小的容量，装价值最大的物品。但是此题 背包容量过大，而价值之和的最大值只有5000，我们可以转换思路，利用01背包的思想，用最大价值来求最小容量；即为，把价值之和看作是背包的容量，进行背包。

代码：

#include
#include
#include
#include
#include
#include
#include
#define max(a,b)(a>b?a:b)
#define min(a,b)(a<b?a:b)
typedef long long ll;
using namespace std;
#define N 5100
ll dp[N],w[N];  ///此时的dp[i]表示的是；价值为i时的最小容量为dp[i];
int v[N];

int main()
{
    int T,i,n,sum;
    ll V;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%I64d",&n,&V);
        sum=0;
        for(i=1;i<=n;i++)
        {
            scanf("%I64d%d",&w[i],&v[i]);
            sum=sum+v[i];
        }

        memset(dp,1000000010,sizeof(dp)); ///要求最小容量，初始化为最大值；
        dp[0]=0;
        for(i=1;i<=n;i++)
        {
            for(int j=sum;j>=v[i];j--)
                dp[j]=min(dp[j],dp[j-v[i]]+w[i]);
        }

        for(i=sum;i>=0;i--)
        {
            if(dp[i]<=V)
            {
               printf("%d\n",i); ///此处输出i，即为满足条件的最大价值
               break;
            }
        }
    }
    return 0;
}