FZU 2214 Knapsack problem (01背包)

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
1
5 15
12 4
2 2
1 1
4 10
1 2
Sample Output
15

题解：01背包，但与标准模型有所不同。weight总量太大，显然不能套标准模型。我们换种思路，动态规划0~sumvalue需要的最小weight。dp初始化为0x3f3f3f3f，dp[0]初始化为0

#include <iostream>#include <cstring>using namespace std;int t, n, b, sv, w[600], v[600], dp[6000];int main() {    scanf("%d", &t);    while (t--) {        scanf("%d%d", &n, &b);        sv = 0;        for (int i = 1; i <= n; ++i) {            scanf("%d%d", &w[i], &v[i]);            sv += v[i];        }        memset(dp, 0x3f, sizeof(dp));        dp[0] = 0;        int ans = 0;        for (int i = 1; i <= n; ++i) {            for (int j = sv; j >= v[i]; --j) {                dp[j] = dp[j - v[i]] + w[i] < dp[j] ? dp[j - v[i]] + w[i] : dp[j];                if (dp[j] <= b && ans < j) ans = j;            }        }        printf("%d\n", ans);    }    return 0;}