B - 真·签到题 FZU - 2214

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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

思路：因为容量太大，所以不能按0-1背包问题去求解。注意到物品个数较小，而且价值和只有5000，所以可以逆向思维，求得对应价值下最小的重量，即dp[i]表示总价值为i是的最小重量是多少，则dp[j] = min(dp[j] , dp[j-val[i]]+vol[i]);最后从v（物品总价值开始判断）开始，找到第一个小于等于w（容量）的v即可。。。。

（我还不太懂，借用一下别人的。。。。）

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[5005];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		while(n--)
		{
			int v[505],w[505];
			memset(dp,0x3f3f3f3f,sizeof(dp));
			int a,b;
			scanf("%d%d",&a,&b);
			int sum=0;
			for(int i=1;i<=a;i++)
			{
				scanf("%d%d",&w[i],&v[i]);
				sum+=v[i];
			}
			dp[0]=0;
			for(int i=1;i<=a;i++)
			{
				for(int j=sum;j>=v[i];j--)
				{
					dp[j]=min(dp[j-v[i]]+w[i],dp[j]);
				}
			}
			for(int i=sum;i>=0;i--)
			{
				if(dp[i]<=b)
				{
					printf("%d\n",i);
					break;
				}
			}
		}
	}
return 0;
}