FZU - 2214 - Knapsack problem（背包 & 思维）

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
题目链接

这个题目也是个01背包的问题，但是问题是，他的质量或说体积实在是太大了，数组是根本开不出来的。所以就感觉有点思维性质的感觉了，我们平时做的01背包都是以重量为下标的，这个时候，题目告诉我们价值和是小于等于500的，所以，我们也可以用价值作为下标。取质量的最小值就可以了。
AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
const int maxn = 5e2 + 5;
const int inf = 0x3f3f3f3f;
int w[maxn], dp[5005];
int v[maxn];

int main()
{
    int T, n, sum, ans;
    int B;
    scanf("%d", &T);
    while(T--)
    {
        sum = 0;
        memset(dp, inf, sizeof(dp));
        dp[0] = 0;
        scanf("%d%d", &n, &B);
        for(int i = 1; i <= n; i++)  scanf("%d%d", &w[i], &v[i]), sum += v[i];
        for(int i = 1; i <= n; 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)
            {
                ans = i;
                break ;
            }
        printf("%d\n", ans);
    }
    return 0;
}